infinity series of $sum_{n=M}^{infty}frac{na^n}{n!}$












2












$begingroup$


Hi I'm self studying calculus, I tried wolframalpha and it says it converges but the final form looks odd, I'm hoping that it converges to a simpler form like what $$sum_{n=0}^{infty}frac{na^n}{n!} = ae^a$$ converges to according to wikipedia. What does $sum_{n=M}^{infty}frac{na^n}{n!}$ converges to if it's offset?






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    It does converge, to what you say it does. I'm confused by your question, what offset are you asking about?
    $endgroup$
    – platty
    Dec 5 '18 at 20:38






  • 1




    $begingroup$
    @platty n begins at M
    $endgroup$
    – drerD
    Dec 5 '18 at 20:40
















2












$begingroup$


Hi I'm self studying calculus, I tried wolframalpha and it says it converges but the final form looks odd, I'm hoping that it converges to a simpler form like what $$sum_{n=0}^{infty}frac{na^n}{n!} = ae^a$$ converges to according to wikipedia. What does $sum_{n=M}^{infty}frac{na^n}{n!}$ converges to if it's offset?






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    It does converge, to what you say it does. I'm confused by your question, what offset are you asking about?
    $endgroup$
    – platty
    Dec 5 '18 at 20:38






  • 1




    $begingroup$
    @platty n begins at M
    $endgroup$
    – drerD
    Dec 5 '18 at 20:40














2












2








2


2



$begingroup$


Hi I'm self studying calculus, I tried wolframalpha and it says it converges but the final form looks odd, I'm hoping that it converges to a simpler form like what $$sum_{n=0}^{infty}frac{na^n}{n!} = ae^a$$ converges to according to wikipedia. What does $sum_{n=M}^{infty}frac{na^n}{n!}$ converges to if it's offset?






sequences-and-series







share|cite|improve this question











$endgroup$




Hi I'm self studying calculus, I tried wolframalpha and it says it converges but the final form looks odd, I'm hoping that it converges to a simpler form like what $$sum_{n=0}^{infty}frac{na^n}{n!} = ae^a$$ converges to according to wikipedia. What does $sum_{n=M}^{infty}frac{na^n}{n!}$ converges to if it's offset?






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 20:40







drerD

















asked Dec 5 '18 at 20:35









drerDdrerD

1559




1559












  • $begingroup$
    It does converge, to what you say it does. I'm confused by your question, what offset are you asking about?
    $endgroup$
    – platty
    Dec 5 '18 at 20:38






  • 1




    $begingroup$
    @platty n begins at M
    $endgroup$
    – drerD
    Dec 5 '18 at 20:40


















  • $begingroup$
    It does converge, to what you say it does. I'm confused by your question, what offset are you asking about?
    $endgroup$
    – platty
    Dec 5 '18 at 20:38






  • 1




    $begingroup$
    @platty n begins at M
    $endgroup$
    – drerD
    Dec 5 '18 at 20:40
















$begingroup$
It does converge, to what you say it does. I'm confused by your question, what offset are you asking about?
$endgroup$
– platty
Dec 5 '18 at 20:38




$begingroup$
It does converge, to what you say it does. I'm confused by your question, what offset are you asking about?
$endgroup$
– platty
Dec 5 '18 at 20:38




1




1




$begingroup$
@platty n begins at M
$endgroup$
– drerD
Dec 5 '18 at 20:40




$begingroup$
@platty n begins at M
$endgroup$
– drerD
Dec 5 '18 at 20:40










3 Answers
3






active

oldest

votes


















2












$begingroup$

It is tied to a special, however analytical function: the Lower Partial Regularized Gamma function $ {gamma (x,a) / Gamma (x)} $
$$
eqalign{
& {{sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{Gamma (x,a)} over {Gamma (x)}} = Q(x,a) cr
& {{sumlimits_{k = x}^infty {{{a^{,k} } over {k!}}} } over {e^{,a} }}
= {{e^{,a} - sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{gamma (x,a)} over {Gamma (x)}} = 1 - Q(x,a) cr}
$$
and $Q(x,a)$ is the Upper Incomplete Regularized Gamma function.



In fact your sum is



$$
eqalign{
& sumlimits_{k = x}^infty {k{{a^{,k} } over {k!}}}
= sumlimits_{k = 0}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = sumlimits_{k = 1}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 1}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = aleft( {sumlimits_{k = 1}^infty {{{a^{,k - 1} } over {left( {k - 1} right)!}}}
- sumlimits_{k = 1}^{x - 1} {{{a^{,k - 1} } over {left( {k - 1} right)!}}} } right) = cr
& = aleft( {sumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 2} {{{a^{,k} } over {k!}}} } right)
= ae^{,a} left( {1 - Q(x - 1,a)} right) cr}
$$



where $x$ is your $M$, but can be also a real or complex number.



-- Addendum --



Let me add some notes to clear how the Truncated Taylor expansion of $e^x$ is connected with the
Gamma function and its corresponding "truncated" versions called Incomplete (lower and upper) Gamma function.

Clearly, the notes here will be just concise hints, to stimulate and address your curiosity to the many treatises available on the subject.



The Gamma function is a $mathbb C to mathbb C$ defined by the functional relation and /or integral transform
$$
eqalign{
& Gamma (z + 1) = z,Gamma (z)quad Rightarrow quad Gamma (n + 1) = n!quad left| {;n in N} right. cr
& Gamma (z) = int_{t, = ,0}^{,infty } {t^{,z - 1} e^{, - t} dt} cr}
$$
(details about domain, convergence, etc. omitted)



The the Lower ($gamma(z,x)$) and the Upper ($Gamma(z,x)$) Incomplete Gamma functions, and
the Regularized version of the Upper Incomplete function ($Q(z,x)$), are defined as
$$
eqalign{
& gamma (z,x) = int_{t, = ,0}^{,x} {t^{,z - 1} e^{, - t} dt} quad Gamma (z,x) = int_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} cr
& gamma (z,x) + Gamma (z,x) = Gamma (z)quad Q(z,x) = {{Gamma (z,x)} over {Gamma (z)}} = 1 - {{gamma (z,x)} over {Gamma (z)}} cr}
$$



Then (very concisely) we have the following chain of relations
$$
eqalign{
& Gamma (1,x) = int_{t, = ,x}^{,infty } {e^{, - t} dt} = e^{, - x} quad Q(0,x) = 0 cr
& Gamma (z + 1,x) = int_{t, = ,x}^{,infty } {t^{,z} e^{, - t} dt}
= - left. {t^{,z} e^{, - t} } right|_{t, = ,x}^{,infty } + zint_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} = cr
& = x^{,z} e^{, - x} + zGamma (z,x) = zGamma (z,x) + x^{,z} Gamma (1,x) cr
& Delta _{,z} Q(z,x) = Q(z + 1,x) - Q(z,x) = {{x^{,z} e^{, - x} } over {Gamma left( {z + 1} right)}} cr
& Q(z,x) = e^{, - x} sumnolimits_{k = 0}^{,z} {{{x^{,k} } over {Gamma left( {k + 1} right)}}} quad quad left| matrix{
;z in mathbb C hfill cr
;x in mathbb C hfill cr} right. cr
& Q(n,x) = e^{, - x} sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} quad left| matrix{
;n in mathbb N hfill cr
;x in mathbb C hfill cr} right. cr}
$$
where
$$
sumnolimits_{k = 0}^{,z} {f(k)}
$$
indicates the Indefinite Sum.



Also refer to this Introduction given at Wolfram Functions Site.



So $Q(n,x)$ is the ratio of the truncated wrt the complete exponential series
$$
Q(n,x) = {{Gamma (n,x)} over {Gamma (n)}} = {{sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} } over {e^{,x} }}
$$
(which of course does not imply that $Gamma(n)=e^x$)



For small integral values of $n$ you just have a simple polynomial $/e^x$.

But if $n$ has large values, or if you want to perform some analysis on that ( in case extending it to non integral values),
then you can take advantage of the expression through the Gamma related functions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I didn't really understand what $Gamma$ was when I saw it in wolframalpha's answer, your answer helps. But why is $Gamma(x)=e^a$ while $Gamma(x,a) = $ the sum? what happens to $Gamma$ when it only takes one parameter x?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:55










  • $begingroup$
    well, I am going to add some more details about in my answer. However, if you are not familiar with the Gamma function, and you actually want to understand what your sum represents (with $M$ integer , but also real or complex), I do suggest that you jump out of W. Alpha and pass and study some papers about .. Gamma (which is a capital function, being the extension of $n!$ to non integer arguments)
    $endgroup$
    – G Cab
    Dec 8 '18 at 15:13



















1












$begingroup$

It converges to



$$ae^a-sum_{n=0}^{M-1}frac{na^n}{n!}$$



for example



$$lim_{Nto+infty}sum_{n=2}^Nfrac{1}{n!}=e-2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I thought about breaking it down but how do you find the closed form of $sum^{M-1}_{n=0}frac{na^n}{n!}$?
    $endgroup$
    – drerD
    Dec 5 '18 at 21:05



















1












$begingroup$

$$sum_{n=M}^{infty}frac{na^n}{n!} = sum_{n=0}^{infty}frac{na^n}{n!}-sum_{n=0}^{M-1}frac{na^n}{n!} = ae^a -sum_{n=0}^{M-1}frac{na^n}{n!}.$$



I can't post a comment, but when you're asking for a closed form for $$sum_{n=0}^{M-1}frac{na^n}{n!}$$



I'm not sure what you mean, since it already is a closed form ($M in mathbb{N}$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    something like this: $sum_{k=0}^{n}x^k = frac{1-x^{n+1}}{1-x}$
    $endgroup$
    – drerD
    Dec 5 '18 at 22:51












  • $begingroup$
    @user14042 I see. It doesn't reduce to something like that. You can use some other named functions or properties but that doesn't achieve anything and complicates it in my opinion.
    $endgroup$
    – Art Vandelay
    Dec 7 '18 at 22:03










  • $begingroup$
    How do we know it's not reducible?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:53











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

It is tied to a special, however analytical function: the Lower Partial Regularized Gamma function $ {gamma (x,a) / Gamma (x)} $
$$
eqalign{
& {{sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{Gamma (x,a)} over {Gamma (x)}} = Q(x,a) cr
& {{sumlimits_{k = x}^infty {{{a^{,k} } over {k!}}} } over {e^{,a} }}
= {{e^{,a} - sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{gamma (x,a)} over {Gamma (x)}} = 1 - Q(x,a) cr}
$$
and $Q(x,a)$ is the Upper Incomplete Regularized Gamma function.



In fact your sum is



$$
eqalign{
& sumlimits_{k = x}^infty {k{{a^{,k} } over {k!}}}
= sumlimits_{k = 0}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = sumlimits_{k = 1}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 1}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = aleft( {sumlimits_{k = 1}^infty {{{a^{,k - 1} } over {left( {k - 1} right)!}}}
- sumlimits_{k = 1}^{x - 1} {{{a^{,k - 1} } over {left( {k - 1} right)!}}} } right) = cr
& = aleft( {sumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 2} {{{a^{,k} } over {k!}}} } right)
= ae^{,a} left( {1 - Q(x - 1,a)} right) cr}
$$



where $x$ is your $M$, but can be also a real or complex number.



-- Addendum --



Let me add some notes to clear how the Truncated Taylor expansion of $e^x$ is connected with the
Gamma function and its corresponding "truncated" versions called Incomplete (lower and upper) Gamma function.

Clearly, the notes here will be just concise hints, to stimulate and address your curiosity to the many treatises available on the subject.



The Gamma function is a $mathbb C to mathbb C$ defined by the functional relation and /or integral transform
$$
eqalign{
& Gamma (z + 1) = z,Gamma (z)quad Rightarrow quad Gamma (n + 1) = n!quad left| {;n in N} right. cr
& Gamma (z) = int_{t, = ,0}^{,infty } {t^{,z - 1} e^{, - t} dt} cr}
$$
(details about domain, convergence, etc. omitted)



The the Lower ($gamma(z,x)$) and the Upper ($Gamma(z,x)$) Incomplete Gamma functions, and
the Regularized version of the Upper Incomplete function ($Q(z,x)$), are defined as
$$
eqalign{
& gamma (z,x) = int_{t, = ,0}^{,x} {t^{,z - 1} e^{, - t} dt} quad Gamma (z,x) = int_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} cr
& gamma (z,x) + Gamma (z,x) = Gamma (z)quad Q(z,x) = {{Gamma (z,x)} over {Gamma (z)}} = 1 - {{gamma (z,x)} over {Gamma (z)}} cr}
$$



Then (very concisely) we have the following chain of relations
$$
eqalign{
& Gamma (1,x) = int_{t, = ,x}^{,infty } {e^{, - t} dt} = e^{, - x} quad Q(0,x) = 0 cr
& Gamma (z + 1,x) = int_{t, = ,x}^{,infty } {t^{,z} e^{, - t} dt}
= - left. {t^{,z} e^{, - t} } right|_{t, = ,x}^{,infty } + zint_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} = cr
& = x^{,z} e^{, - x} + zGamma (z,x) = zGamma (z,x) + x^{,z} Gamma (1,x) cr
& Delta _{,z} Q(z,x) = Q(z + 1,x) - Q(z,x) = {{x^{,z} e^{, - x} } over {Gamma left( {z + 1} right)}} cr
& Q(z,x) = e^{, - x} sumnolimits_{k = 0}^{,z} {{{x^{,k} } over {Gamma left( {k + 1} right)}}} quad quad left| matrix{
;z in mathbb C hfill cr
;x in mathbb C hfill cr} right. cr
& Q(n,x) = e^{, - x} sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} quad left| matrix{
;n in mathbb N hfill cr
;x in mathbb C hfill cr} right. cr}
$$
where
$$
sumnolimits_{k = 0}^{,z} {f(k)}
$$
indicates the Indefinite Sum.



Also refer to this Introduction given at Wolfram Functions Site.



So $Q(n,x)$ is the ratio of the truncated wrt the complete exponential series
$$
Q(n,x) = {{Gamma (n,x)} over {Gamma (n)}} = {{sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} } over {e^{,x} }}
$$
(which of course does not imply that $Gamma(n)=e^x$)



For small integral values of $n$ you just have a simple polynomial $/e^x$.

But if $n$ has large values, or if you want to perform some analysis on that ( in case extending it to non integral values),
then you can take advantage of the expression through the Gamma related functions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I didn't really understand what $Gamma$ was when I saw it in wolframalpha's answer, your answer helps. But why is $Gamma(x)=e^a$ while $Gamma(x,a) = $ the sum? what happens to $Gamma$ when it only takes one parameter x?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:55










  • $begingroup$
    well, I am going to add some more details about in my answer. However, if you are not familiar with the Gamma function, and you actually want to understand what your sum represents (with $M$ integer , but also real or complex), I do suggest that you jump out of W. Alpha and pass and study some papers about .. Gamma (which is a capital function, being the extension of $n!$ to non integer arguments)
    $endgroup$
    – G Cab
    Dec 8 '18 at 15:13
















2












$begingroup$

It is tied to a special, however analytical function: the Lower Partial Regularized Gamma function $ {gamma (x,a) / Gamma (x)} $
$$
eqalign{
& {{sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{Gamma (x,a)} over {Gamma (x)}} = Q(x,a) cr
& {{sumlimits_{k = x}^infty {{{a^{,k} } over {k!}}} } over {e^{,a} }}
= {{e^{,a} - sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{gamma (x,a)} over {Gamma (x)}} = 1 - Q(x,a) cr}
$$
and $Q(x,a)$ is the Upper Incomplete Regularized Gamma function.



In fact your sum is



$$
eqalign{
& sumlimits_{k = x}^infty {k{{a^{,k} } over {k!}}}
= sumlimits_{k = 0}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = sumlimits_{k = 1}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 1}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = aleft( {sumlimits_{k = 1}^infty {{{a^{,k - 1} } over {left( {k - 1} right)!}}}
- sumlimits_{k = 1}^{x - 1} {{{a^{,k - 1} } over {left( {k - 1} right)!}}} } right) = cr
& = aleft( {sumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 2} {{{a^{,k} } over {k!}}} } right)
= ae^{,a} left( {1 - Q(x - 1,a)} right) cr}
$$



where $x$ is your $M$, but can be also a real or complex number.



-- Addendum --



Let me add some notes to clear how the Truncated Taylor expansion of $e^x$ is connected with the
Gamma function and its corresponding "truncated" versions called Incomplete (lower and upper) Gamma function.

Clearly, the notes here will be just concise hints, to stimulate and address your curiosity to the many treatises available on the subject.



The Gamma function is a $mathbb C to mathbb C$ defined by the functional relation and /or integral transform
$$
eqalign{
& Gamma (z + 1) = z,Gamma (z)quad Rightarrow quad Gamma (n + 1) = n!quad left| {;n in N} right. cr
& Gamma (z) = int_{t, = ,0}^{,infty } {t^{,z - 1} e^{, - t} dt} cr}
$$
(details about domain, convergence, etc. omitted)



The the Lower ($gamma(z,x)$) and the Upper ($Gamma(z,x)$) Incomplete Gamma functions, and
the Regularized version of the Upper Incomplete function ($Q(z,x)$), are defined as
$$
eqalign{
& gamma (z,x) = int_{t, = ,0}^{,x} {t^{,z - 1} e^{, - t} dt} quad Gamma (z,x) = int_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} cr
& gamma (z,x) + Gamma (z,x) = Gamma (z)quad Q(z,x) = {{Gamma (z,x)} over {Gamma (z)}} = 1 - {{gamma (z,x)} over {Gamma (z)}} cr}
$$



Then (very concisely) we have the following chain of relations
$$
eqalign{
& Gamma (1,x) = int_{t, = ,x}^{,infty } {e^{, - t} dt} = e^{, - x} quad Q(0,x) = 0 cr
& Gamma (z + 1,x) = int_{t, = ,x}^{,infty } {t^{,z} e^{, - t} dt}
= - left. {t^{,z} e^{, - t} } right|_{t, = ,x}^{,infty } + zint_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} = cr
& = x^{,z} e^{, - x} + zGamma (z,x) = zGamma (z,x) + x^{,z} Gamma (1,x) cr
& Delta _{,z} Q(z,x) = Q(z + 1,x) - Q(z,x) = {{x^{,z} e^{, - x} } over {Gamma left( {z + 1} right)}} cr
& Q(z,x) = e^{, - x} sumnolimits_{k = 0}^{,z} {{{x^{,k} } over {Gamma left( {k + 1} right)}}} quad quad left| matrix{
;z in mathbb C hfill cr
;x in mathbb C hfill cr} right. cr
& Q(n,x) = e^{, - x} sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} quad left| matrix{
;n in mathbb N hfill cr
;x in mathbb C hfill cr} right. cr}
$$
where
$$
sumnolimits_{k = 0}^{,z} {f(k)}
$$
indicates the Indefinite Sum.



Also refer to this Introduction given at Wolfram Functions Site.



So $Q(n,x)$ is the ratio of the truncated wrt the complete exponential series
$$
Q(n,x) = {{Gamma (n,x)} over {Gamma (n)}} = {{sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} } over {e^{,x} }}
$$
(which of course does not imply that $Gamma(n)=e^x$)



For small integral values of $n$ you just have a simple polynomial $/e^x$.

But if $n$ has large values, or if you want to perform some analysis on that ( in case extending it to non integral values),
then you can take advantage of the expression through the Gamma related functions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I didn't really understand what $Gamma$ was when I saw it in wolframalpha's answer, your answer helps. But why is $Gamma(x)=e^a$ while $Gamma(x,a) = $ the sum? what happens to $Gamma$ when it only takes one parameter x?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:55










  • $begingroup$
    well, I am going to add some more details about in my answer. However, if you are not familiar with the Gamma function, and you actually want to understand what your sum represents (with $M$ integer , but also real or complex), I do suggest that you jump out of W. Alpha and pass and study some papers about .. Gamma (which is a capital function, being the extension of $n!$ to non integer arguments)
    $endgroup$
    – G Cab
    Dec 8 '18 at 15:13














2












2








2





$begingroup$

It is tied to a special, however analytical function: the Lower Partial Regularized Gamma function $ {gamma (x,a) / Gamma (x)} $
$$
eqalign{
& {{sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{Gamma (x,a)} over {Gamma (x)}} = Q(x,a) cr
& {{sumlimits_{k = x}^infty {{{a^{,k} } over {k!}}} } over {e^{,a} }}
= {{e^{,a} - sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{gamma (x,a)} over {Gamma (x)}} = 1 - Q(x,a) cr}
$$
and $Q(x,a)$ is the Upper Incomplete Regularized Gamma function.



In fact your sum is



$$
eqalign{
& sumlimits_{k = x}^infty {k{{a^{,k} } over {k!}}}
= sumlimits_{k = 0}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = sumlimits_{k = 1}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 1}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = aleft( {sumlimits_{k = 1}^infty {{{a^{,k - 1} } over {left( {k - 1} right)!}}}
- sumlimits_{k = 1}^{x - 1} {{{a^{,k - 1} } over {left( {k - 1} right)!}}} } right) = cr
& = aleft( {sumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 2} {{{a^{,k} } over {k!}}} } right)
= ae^{,a} left( {1 - Q(x - 1,a)} right) cr}
$$



where $x$ is your $M$, but can be also a real or complex number.



-- Addendum --



Let me add some notes to clear how the Truncated Taylor expansion of $e^x$ is connected with the
Gamma function and its corresponding "truncated" versions called Incomplete (lower and upper) Gamma function.

Clearly, the notes here will be just concise hints, to stimulate and address your curiosity to the many treatises available on the subject.



The Gamma function is a $mathbb C to mathbb C$ defined by the functional relation and /or integral transform
$$
eqalign{
& Gamma (z + 1) = z,Gamma (z)quad Rightarrow quad Gamma (n + 1) = n!quad left| {;n in N} right. cr
& Gamma (z) = int_{t, = ,0}^{,infty } {t^{,z - 1} e^{, - t} dt} cr}
$$
(details about domain, convergence, etc. omitted)



The the Lower ($gamma(z,x)$) and the Upper ($Gamma(z,x)$) Incomplete Gamma functions, and
the Regularized version of the Upper Incomplete function ($Q(z,x)$), are defined as
$$
eqalign{
& gamma (z,x) = int_{t, = ,0}^{,x} {t^{,z - 1} e^{, - t} dt} quad Gamma (z,x) = int_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} cr
& gamma (z,x) + Gamma (z,x) = Gamma (z)quad Q(z,x) = {{Gamma (z,x)} over {Gamma (z)}} = 1 - {{gamma (z,x)} over {Gamma (z)}} cr}
$$



Then (very concisely) we have the following chain of relations
$$
eqalign{
& Gamma (1,x) = int_{t, = ,x}^{,infty } {e^{, - t} dt} = e^{, - x} quad Q(0,x) = 0 cr
& Gamma (z + 1,x) = int_{t, = ,x}^{,infty } {t^{,z} e^{, - t} dt}
= - left. {t^{,z} e^{, - t} } right|_{t, = ,x}^{,infty } + zint_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} = cr
& = x^{,z} e^{, - x} + zGamma (z,x) = zGamma (z,x) + x^{,z} Gamma (1,x) cr
& Delta _{,z} Q(z,x) = Q(z + 1,x) - Q(z,x) = {{x^{,z} e^{, - x} } over {Gamma left( {z + 1} right)}} cr
& Q(z,x) = e^{, - x} sumnolimits_{k = 0}^{,z} {{{x^{,k} } over {Gamma left( {k + 1} right)}}} quad quad left| matrix{
;z in mathbb C hfill cr
;x in mathbb C hfill cr} right. cr
& Q(n,x) = e^{, - x} sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} quad left| matrix{
;n in mathbb N hfill cr
;x in mathbb C hfill cr} right. cr}
$$
where
$$
sumnolimits_{k = 0}^{,z} {f(k)}
$$
indicates the Indefinite Sum.



Also refer to this Introduction given at Wolfram Functions Site.



So $Q(n,x)$ is the ratio of the truncated wrt the complete exponential series
$$
Q(n,x) = {{Gamma (n,x)} over {Gamma (n)}} = {{sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} } over {e^{,x} }}
$$
(which of course does not imply that $Gamma(n)=e^x$)



For small integral values of $n$ you just have a simple polynomial $/e^x$.

But if $n$ has large values, or if you want to perform some analysis on that ( in case extending it to non integral values),
then you can take advantage of the expression through the Gamma related functions.






share|cite|improve this answer











$endgroup$



It is tied to a special, however analytical function: the Lower Partial Regularized Gamma function $ {gamma (x,a) / Gamma (x)} $
$$
eqalign{
& {{sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{Gamma (x,a)} over {Gamma (x)}} = Q(x,a) cr
& {{sumlimits_{k = x}^infty {{{a^{,k} } over {k!}}} } over {e^{,a} }}
= {{e^{,a} - sumlimits_{k = 0}^{x - 1} {{{a^{,k} } over {k!}}} } over {e^{,a} }} = {{gamma (x,a)} over {Gamma (x)}} = 1 - Q(x,a) cr}
$$
and $Q(x,a)$ is the Upper Incomplete Regularized Gamma function.



In fact your sum is



$$
eqalign{
& sumlimits_{k = x}^infty {k{{a^{,k} } over {k!}}}
= sumlimits_{k = 0}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = sumlimits_{k = 1}^infty {k{{a^{,k} } over {k!}}} - sumlimits_{k = 1}^{x - 1} {k{{a^{,k} } over {k!}}} = cr
& = aleft( {sumlimits_{k = 1}^infty {{{a^{,k - 1} } over {left( {k - 1} right)!}}}
- sumlimits_{k = 1}^{x - 1} {{{a^{,k - 1} } over {left( {k - 1} right)!}}} } right) = cr
& = aleft( {sumlimits_{k = 0}^infty {{{a^{,k} } over {k!}}} - sumlimits_{k = 0}^{x - 2} {{{a^{,k} } over {k!}}} } right)
= ae^{,a} left( {1 - Q(x - 1,a)} right) cr}
$$



where $x$ is your $M$, but can be also a real or complex number.



-- Addendum --



Let me add some notes to clear how the Truncated Taylor expansion of $e^x$ is connected with the
Gamma function and its corresponding "truncated" versions called Incomplete (lower and upper) Gamma function.

Clearly, the notes here will be just concise hints, to stimulate and address your curiosity to the many treatises available on the subject.



The Gamma function is a $mathbb C to mathbb C$ defined by the functional relation and /or integral transform
$$
eqalign{
& Gamma (z + 1) = z,Gamma (z)quad Rightarrow quad Gamma (n + 1) = n!quad left| {;n in N} right. cr
& Gamma (z) = int_{t, = ,0}^{,infty } {t^{,z - 1} e^{, - t} dt} cr}
$$
(details about domain, convergence, etc. omitted)



The the Lower ($gamma(z,x)$) and the Upper ($Gamma(z,x)$) Incomplete Gamma functions, and
the Regularized version of the Upper Incomplete function ($Q(z,x)$), are defined as
$$
eqalign{
& gamma (z,x) = int_{t, = ,0}^{,x} {t^{,z - 1} e^{, - t} dt} quad Gamma (z,x) = int_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} cr
& gamma (z,x) + Gamma (z,x) = Gamma (z)quad Q(z,x) = {{Gamma (z,x)} over {Gamma (z)}} = 1 - {{gamma (z,x)} over {Gamma (z)}} cr}
$$



Then (very concisely) we have the following chain of relations
$$
eqalign{
& Gamma (1,x) = int_{t, = ,x}^{,infty } {e^{, - t} dt} = e^{, - x} quad Q(0,x) = 0 cr
& Gamma (z + 1,x) = int_{t, = ,x}^{,infty } {t^{,z} e^{, - t} dt}
= - left. {t^{,z} e^{, - t} } right|_{t, = ,x}^{,infty } + zint_{t, = ,x}^{,infty } {t^{,z - 1} e^{, - t} dt} = cr
& = x^{,z} e^{, - x} + zGamma (z,x) = zGamma (z,x) + x^{,z} Gamma (1,x) cr
& Delta _{,z} Q(z,x) = Q(z + 1,x) - Q(z,x) = {{x^{,z} e^{, - x} } over {Gamma left( {z + 1} right)}} cr
& Q(z,x) = e^{, - x} sumnolimits_{k = 0}^{,z} {{{x^{,k} } over {Gamma left( {k + 1} right)}}} quad quad left| matrix{
;z in mathbb C hfill cr
;x in mathbb C hfill cr} right. cr
& Q(n,x) = e^{, - x} sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} quad left| matrix{
;n in mathbb N hfill cr
;x in mathbb C hfill cr} right. cr}
$$
where
$$
sumnolimits_{k = 0}^{,z} {f(k)}
$$
indicates the Indefinite Sum.



Also refer to this Introduction given at Wolfram Functions Site.



So $Q(n,x)$ is the ratio of the truncated wrt the complete exponential series
$$
Q(n,x) = {{Gamma (n,x)} over {Gamma (n)}} = {{sumlimits_{k = 0}^{n - 1} {{{x^{,k} } over {k!}}} } over {e^{,x} }}
$$
(which of course does not imply that $Gamma(n)=e^x$)



For small integral values of $n$ you just have a simple polynomial $/e^x$.

But if $n$ has large values, or if you want to perform some analysis on that ( in case extending it to non integral values),
then you can take advantage of the expression through the Gamma related functions.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 18:13

























answered Dec 5 '18 at 22:40









G CabG Cab

18.5k31237




18.5k31237












  • $begingroup$
    I didn't really understand what $Gamma$ was when I saw it in wolframalpha's answer, your answer helps. But why is $Gamma(x)=e^a$ while $Gamma(x,a) = $ the sum? what happens to $Gamma$ when it only takes one parameter x?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:55










  • $begingroup$
    well, I am going to add some more details about in my answer. However, if you are not familiar with the Gamma function, and you actually want to understand what your sum represents (with $M$ integer , but also real or complex), I do suggest that you jump out of W. Alpha and pass and study some papers about .. Gamma (which is a capital function, being the extension of $n!$ to non integer arguments)
    $endgroup$
    – G Cab
    Dec 8 '18 at 15:13


















  • $begingroup$
    I didn't really understand what $Gamma$ was when I saw it in wolframalpha's answer, your answer helps. But why is $Gamma(x)=e^a$ while $Gamma(x,a) = $ the sum? what happens to $Gamma$ when it only takes one parameter x?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:55










  • $begingroup$
    well, I am going to add some more details about in my answer. However, if you are not familiar with the Gamma function, and you actually want to understand what your sum represents (with $M$ integer , but also real or complex), I do suggest that you jump out of W. Alpha and pass and study some papers about .. Gamma (which is a capital function, being the extension of $n!$ to non integer arguments)
    $endgroup$
    – G Cab
    Dec 8 '18 at 15:13
















$begingroup$
I didn't really understand what $Gamma$ was when I saw it in wolframalpha's answer, your answer helps. But why is $Gamma(x)=e^a$ while $Gamma(x,a) = $ the sum? what happens to $Gamma$ when it only takes one parameter x?
$endgroup$
– drerD
Dec 8 '18 at 0:55




$begingroup$
I didn't really understand what $Gamma$ was when I saw it in wolframalpha's answer, your answer helps. But why is $Gamma(x)=e^a$ while $Gamma(x,a) = $ the sum? what happens to $Gamma$ when it only takes one parameter x?
$endgroup$
– drerD
Dec 8 '18 at 0:55












$begingroup$
well, I am going to add some more details about in my answer. However, if you are not familiar with the Gamma function, and you actually want to understand what your sum represents (with $M$ integer , but also real or complex), I do suggest that you jump out of W. Alpha and pass and study some papers about .. Gamma (which is a capital function, being the extension of $n!$ to non integer arguments)
$endgroup$
– G Cab
Dec 8 '18 at 15:13




$begingroup$
well, I am going to add some more details about in my answer. However, if you are not familiar with the Gamma function, and you actually want to understand what your sum represents (with $M$ integer , but also real or complex), I do suggest that you jump out of W. Alpha and pass and study some papers about .. Gamma (which is a capital function, being the extension of $n!$ to non integer arguments)
$endgroup$
– G Cab
Dec 8 '18 at 15:13











1












$begingroup$

It converges to



$$ae^a-sum_{n=0}^{M-1}frac{na^n}{n!}$$



for example



$$lim_{Nto+infty}sum_{n=2}^Nfrac{1}{n!}=e-2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I thought about breaking it down but how do you find the closed form of $sum^{M-1}_{n=0}frac{na^n}{n!}$?
    $endgroup$
    – drerD
    Dec 5 '18 at 21:05
















1












$begingroup$

It converges to



$$ae^a-sum_{n=0}^{M-1}frac{na^n}{n!}$$



for example



$$lim_{Nto+infty}sum_{n=2}^Nfrac{1}{n!}=e-2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I thought about breaking it down but how do you find the closed form of $sum^{M-1}_{n=0}frac{na^n}{n!}$?
    $endgroup$
    – drerD
    Dec 5 '18 at 21:05














1












1








1





$begingroup$

It converges to



$$ae^a-sum_{n=0}^{M-1}frac{na^n}{n!}$$



for example



$$lim_{Nto+infty}sum_{n=2}^Nfrac{1}{n!}=e-2$$






share|cite|improve this answer









$endgroup$



It converges to



$$ae^a-sum_{n=0}^{M-1}frac{na^n}{n!}$$



for example



$$lim_{Nto+infty}sum_{n=2}^Nfrac{1}{n!}=e-2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 20:46









hamam_Abdallahhamam_Abdallah

38k21634




38k21634












  • $begingroup$
    I thought about breaking it down but how do you find the closed form of $sum^{M-1}_{n=0}frac{na^n}{n!}$?
    $endgroup$
    – drerD
    Dec 5 '18 at 21:05


















  • $begingroup$
    I thought about breaking it down but how do you find the closed form of $sum^{M-1}_{n=0}frac{na^n}{n!}$?
    $endgroup$
    – drerD
    Dec 5 '18 at 21:05
















$begingroup$
I thought about breaking it down but how do you find the closed form of $sum^{M-1}_{n=0}frac{na^n}{n!}$?
$endgroup$
– drerD
Dec 5 '18 at 21:05




$begingroup$
I thought about breaking it down but how do you find the closed form of $sum^{M-1}_{n=0}frac{na^n}{n!}$?
$endgroup$
– drerD
Dec 5 '18 at 21:05











1












$begingroup$

$$sum_{n=M}^{infty}frac{na^n}{n!} = sum_{n=0}^{infty}frac{na^n}{n!}-sum_{n=0}^{M-1}frac{na^n}{n!} = ae^a -sum_{n=0}^{M-1}frac{na^n}{n!}.$$



I can't post a comment, but when you're asking for a closed form for $$sum_{n=0}^{M-1}frac{na^n}{n!}$$



I'm not sure what you mean, since it already is a closed form ($M in mathbb{N}$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    something like this: $sum_{k=0}^{n}x^k = frac{1-x^{n+1}}{1-x}$
    $endgroup$
    – drerD
    Dec 5 '18 at 22:51












  • $begingroup$
    @user14042 I see. It doesn't reduce to something like that. You can use some other named functions or properties but that doesn't achieve anything and complicates it in my opinion.
    $endgroup$
    – Art Vandelay
    Dec 7 '18 at 22:03










  • $begingroup$
    How do we know it's not reducible?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:53
















1












$begingroup$

$$sum_{n=M}^{infty}frac{na^n}{n!} = sum_{n=0}^{infty}frac{na^n}{n!}-sum_{n=0}^{M-1}frac{na^n}{n!} = ae^a -sum_{n=0}^{M-1}frac{na^n}{n!}.$$



I can't post a comment, but when you're asking for a closed form for $$sum_{n=0}^{M-1}frac{na^n}{n!}$$



I'm not sure what you mean, since it already is a closed form ($M in mathbb{N}$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    something like this: $sum_{k=0}^{n}x^k = frac{1-x^{n+1}}{1-x}$
    $endgroup$
    – drerD
    Dec 5 '18 at 22:51












  • $begingroup$
    @user14042 I see. It doesn't reduce to something like that. You can use some other named functions or properties but that doesn't achieve anything and complicates it in my opinion.
    $endgroup$
    – Art Vandelay
    Dec 7 '18 at 22:03










  • $begingroup$
    How do we know it's not reducible?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:53














1












1








1





$begingroup$

$$sum_{n=M}^{infty}frac{na^n}{n!} = sum_{n=0}^{infty}frac{na^n}{n!}-sum_{n=0}^{M-1}frac{na^n}{n!} = ae^a -sum_{n=0}^{M-1}frac{na^n}{n!}.$$



I can't post a comment, but when you're asking for a closed form for $$sum_{n=0}^{M-1}frac{na^n}{n!}$$



I'm not sure what you mean, since it already is a closed form ($M in mathbb{N}$).






share|cite|improve this answer









$endgroup$



$$sum_{n=M}^{infty}frac{na^n}{n!} = sum_{n=0}^{infty}frac{na^n}{n!}-sum_{n=0}^{M-1}frac{na^n}{n!} = ae^a -sum_{n=0}^{M-1}frac{na^n}{n!}.$$



I can't post a comment, but when you're asking for a closed form for $$sum_{n=0}^{M-1}frac{na^n}{n!}$$



I'm not sure what you mean, since it already is a closed form ($M in mathbb{N}$).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 22:21









Art VandelayArt Vandelay

133




133












  • $begingroup$
    something like this: $sum_{k=0}^{n}x^k = frac{1-x^{n+1}}{1-x}$
    $endgroup$
    – drerD
    Dec 5 '18 at 22:51












  • $begingroup$
    @user14042 I see. It doesn't reduce to something like that. You can use some other named functions or properties but that doesn't achieve anything and complicates it in my opinion.
    $endgroup$
    – Art Vandelay
    Dec 7 '18 at 22:03










  • $begingroup$
    How do we know it's not reducible?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:53


















  • $begingroup$
    something like this: $sum_{k=0}^{n}x^k = frac{1-x^{n+1}}{1-x}$
    $endgroup$
    – drerD
    Dec 5 '18 at 22:51












  • $begingroup$
    @user14042 I see. It doesn't reduce to something like that. You can use some other named functions or properties but that doesn't achieve anything and complicates it in my opinion.
    $endgroup$
    – Art Vandelay
    Dec 7 '18 at 22:03










  • $begingroup$
    How do we know it's not reducible?
    $endgroup$
    – drerD
    Dec 8 '18 at 0:53
















$begingroup$
something like this: $sum_{k=0}^{n}x^k = frac{1-x^{n+1}}{1-x}$
$endgroup$
– drerD
Dec 5 '18 at 22:51






$begingroup$
something like this: $sum_{k=0}^{n}x^k = frac{1-x^{n+1}}{1-x}$
$endgroup$
– drerD
Dec 5 '18 at 22:51














$begingroup$
@user14042 I see. It doesn't reduce to something like that. You can use some other named functions or properties but that doesn't achieve anything and complicates it in my opinion.
$endgroup$
– Art Vandelay
Dec 7 '18 at 22:03




$begingroup$
@user14042 I see. It doesn't reduce to something like that. You can use some other named functions or properties but that doesn't achieve anything and complicates it in my opinion.
$endgroup$
– Art Vandelay
Dec 7 '18 at 22:03












$begingroup$
How do we know it's not reducible?
$endgroup$
– drerD
Dec 8 '18 at 0:53




$begingroup$
How do we know it's not reducible?
$endgroup$
– drerD
Dec 8 '18 at 0:53


















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