Two sums, the relation between them
What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.
$$sumlimits_{n=2}^{infty}(frac{1}{n}-frac{1}{n-1})=-1$$
and
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=frac{pi^2}{6}-1$$
real-analysis integration sequences-and-series number-theory power-series
add a comment |
What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.
$$sumlimits_{n=2}^{infty}(frac{1}{n}-frac{1}{n-1})=-1$$
and
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=frac{pi^2}{6}-1$$
real-analysis integration sequences-and-series number-theory power-series
The second sum looks wrong. For one thing, the value should be negative...
– Erick Wong
Oct 19 '18 at 2:42
add a comment |
What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.
$$sumlimits_{n=2}^{infty}(frac{1}{n}-frac{1}{n-1})=-1$$
and
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=frac{pi^2}{6}-1$$
real-analysis integration sequences-and-series number-theory power-series
What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.
$$sumlimits_{n=2}^{infty}(frac{1}{n}-frac{1}{n-1})=-1$$
and
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=frac{pi^2}{6}-1$$
real-analysis integration sequences-and-series number-theory power-series
real-analysis integration sequences-and-series number-theory power-series
edited Oct 19 '18 at 12:20
Marianna Kalwat
asked Oct 19 '18 at 1:55
Marianna KalwatMarianna Kalwat
626
626
The second sum looks wrong. For one thing, the value should be negative...
– Erick Wong
Oct 19 '18 at 2:42
add a comment |
The second sum looks wrong. For one thing, the value should be negative...
– Erick Wong
Oct 19 '18 at 2:42
The second sum looks wrong. For one thing, the value should be negative...
– Erick Wong
Oct 19 '18 at 2:42
The second sum looks wrong. For one thing, the value should be negative...
– Erick Wong
Oct 19 '18 at 2:42
add a comment |
2 Answers
2
active
oldest
votes
I do not understand much what you mean but maybe these points help you in something
1)The last result is wrong, first think that you can separate this sum into two parts
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$
then you can see that
$$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$
and at the same time that
$$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$
then using what has already been said and this sum ,discovered by euler,$
sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$ We can replace it and gives us
$$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$
2)Also in the same way we can generalize these sums for any exponent
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$
Applying the same steps as before we get
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$
Thanks you for reading.
How you get $ zeta(k)-1-zeta(k)$?
– Marianna Kalwat
Oct 19 '18 at 12:26
we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
– Blas Casazza
Oct 19 '18 at 17:13
add a comment |
In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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I do not understand much what you mean but maybe these points help you in something
1)The last result is wrong, first think that you can separate this sum into two parts
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$
then you can see that
$$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$
and at the same time that
$$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$
then using what has already been said and this sum ,discovered by euler,$
sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$ We can replace it and gives us
$$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$
2)Also in the same way we can generalize these sums for any exponent
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$
Applying the same steps as before we get
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$
Thanks you for reading.
How you get $ zeta(k)-1-zeta(k)$?
– Marianna Kalwat
Oct 19 '18 at 12:26
we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
– Blas Casazza
Oct 19 '18 at 17:13
add a comment |
I do not understand much what you mean but maybe these points help you in something
1)The last result is wrong, first think that you can separate this sum into two parts
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$
then you can see that
$$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$
and at the same time that
$$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$
then using what has already been said and this sum ,discovered by euler,$
sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$ We can replace it and gives us
$$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$
2)Also in the same way we can generalize these sums for any exponent
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$
Applying the same steps as before we get
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$
Thanks you for reading.
How you get $ zeta(k)-1-zeta(k)$?
– Marianna Kalwat
Oct 19 '18 at 12:26
we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
– Blas Casazza
Oct 19 '18 at 17:13
add a comment |
I do not understand much what you mean but maybe these points help you in something
1)The last result is wrong, first think that you can separate this sum into two parts
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$
then you can see that
$$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$
and at the same time that
$$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$
then using what has already been said and this sum ,discovered by euler,$
sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$ We can replace it and gives us
$$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$
2)Also in the same way we can generalize these sums for any exponent
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$
Applying the same steps as before we get
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$
Thanks you for reading.
I do not understand much what you mean but maybe these points help you in something
1)The last result is wrong, first think that you can separate this sum into two parts
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}$$
then you can see that
$$sumlimits_{n=2}^{infty}frac{1}{n^2}=sumlimits_{n=1}^{infty}frac{1}{n^2}-(frac{1}{(n=1)^2})=sumlimits_{n=1}^{infty}frac{1}{n^2}-1$$
and at the same time that
$$sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=sumlimits_{n=1}^{infty}frac{1}{n^2}$$
then using what has already been said and this sum ,discovered by euler,$
sumlimits_{n=1}^{infty}frac{1}{n^2}=frac{pi^2}{6}$ We can replace it and gives us
$$sumlimits_{n=2}^{infty}frac{1}{n^2}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^2}}=frac{pi^2}{6}-1-frac{pi^2}{6}=-1$$
2)Also in the same way we can generalize these sums for any exponent
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})$$
Applying the same steps as before we get
$$sumlimits_{n=2}^{infty}(frac{1}{n^k}-frac{1}{(n-1)^k})=sumlimits_{n=2}^{infty}frac{1}{n^k}-sumlimits_{n=2}^{infty}{frac{1}{(n-1)^k}}=zeta(k)-1-zeta(k)=-1$$
Thanks you for reading.
answered Oct 19 '18 at 6:20
Blas CasazzaBlas Casazza
527
527
How you get $ zeta(k)-1-zeta(k)$?
– Marianna Kalwat
Oct 19 '18 at 12:26
we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
– Blas Casazza
Oct 19 '18 at 17:13
add a comment |
How you get $ zeta(k)-1-zeta(k)$?
– Marianna Kalwat
Oct 19 '18 at 12:26
we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
– Blas Casazza
Oct 19 '18 at 17:13
How you get $ zeta(k)-1-zeta(k)$?
– Marianna Kalwat
Oct 19 '18 at 12:26
How you get $ zeta(k)-1-zeta(k)$?
– Marianna Kalwat
Oct 19 '18 at 12:26
we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
– Blas Casazza
Oct 19 '18 at 17:13
we have to do the same steps, keep in mind that $sumlimits_{n=2}^{infty}frac{1}{n^k}=sumlimits_{n=1}^{infty}frac{1}{n^k}-frac{1}{({n^k=1})}=zeta(k)-1$ and at the same time $sumlimits_{n=2}^{infty}frac{1}{(n-1)^k}=zeta(k)$
– Blas Casazza
Oct 19 '18 at 17:13
add a comment |
In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$
add a comment |
In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$
add a comment |
In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$
In fact we have $$sum_{n=1}^{infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+cdots=a_{infty}-a_1$$
These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.
$$sumlimits_{n=2}^{infty}(frac{1}{n^2}-frac{1}{(n-1)^2})=-1$$
answered Nov 30 '18 at 19:20
Mostafa AyazMostafa Ayaz
14.5k3937
14.5k3937
add a comment |
add a comment |
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The second sum looks wrong. For one thing, the value should be negative...
– Erick Wong
Oct 19 '18 at 2:42