Different ways of proving that $|sum^{infty}_{k=1}(1-cos(1/k))|leq 2 $












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I've found two ways of proving that
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&leq 2 end{align}
Are there any other ways out there, for proving this?



METHOD 1



Let $kin Bbb{N}$, then



begin{align} f:[ 0&,1]longrightarrow Bbb{R}\&x mapsto
cosleft(dfrac{x}{k}right) end{align}

is continuous. Then, by Mean Value Theorem, there exists $cin [ 0,x]$ such that
begin{align} cosleft(dfrac{x}{k}right)-cosleft(0right) =-dfrac{1}{k}sinleft(dfrac{c}{k}right),(x-0), end{align}
which implies begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right| &=left|sum^{infty}_{k=1}dfrac{x}{k}sinleft(dfrac{c}{k}right)right| leq sum^{infty}_{k=1}dfrac{left|xright|}{k}left|sinleft(dfrac{c}{k}right)right|leq sum^{infty}_{k=1}dfrac{left|xright|}{k}dfrac{left|cright|}{k}\&leq sum^{infty}_{k=1}left(dfrac{left|xright|}{k}right)^2leq sum^{infty}_{k=1}dfrac{1}{k^2}=1+ sum^{infty}_{k=2}dfrac{1}{k^2}\&leq 1+ sum^{infty}_{k=2}dfrac{1}{k(k-1)}\&= 1+ limlimits_{ntoinfty}sum^{n}_{k=2}left(dfrac{1}{k-1}-dfrac{1}{k}right)\&=1+ limlimits_{ntoinfty}left(1-dfrac{1}{n}right)\&=2, end{align}



METHOD 2



Let $xin [0,1]$ be fixed, then
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}dfrac{1}{k}left[-kcosleft(dfrac{x}{k}right)right]^{1}_{0}=sum^{infty}_{k=1}dfrac{1}{k}int^{1}_{0}sinleft(dfrac{x}{k}right)dx \&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx end{align}
The series $sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)$ converges uniformly on $[0,1]$, by Weierstrass-M test, since
begin{align} left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k^2}. end{align}
Hence,
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx=int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx, end{align}
and
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&=left|int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dxright|leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right)right|dx \&leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k^2}dx \&leq 2 end{align}










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  • $begingroup$
    Nice work! (+1) Do you know if the series $$sum_{ngeq1}1-cos(1/k)$$ Has a closed form?
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:14










  • $begingroup$
    Hmm... I don't know but do you have an idea its closed form?
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:16












  • $begingroup$
    I do not. It looks a lot like a Fourier series though. Special functions may help...
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:21












  • $begingroup$
    @clathratus: Kindly see Robert Israel's answer for a closed form.
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:35










  • $begingroup$
    Hi @Mike ! I would like to learn how to evaluate to solve some problems like the one you posted, but I've never seen such problems in my Analysis books. Could you please share how you learned to solve them and if there is any resource with many problems like this, either teaching you how to do them or just many practice problems?
    $endgroup$
    – Ovi
    Jan 1 at 0:36
















7












$begingroup$


I've found two ways of proving that
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&leq 2 end{align}
Are there any other ways out there, for proving this?



METHOD 1



Let $kin Bbb{N}$, then



begin{align} f:[ 0&,1]longrightarrow Bbb{R}\&x mapsto
cosleft(dfrac{x}{k}right) end{align}

is continuous. Then, by Mean Value Theorem, there exists $cin [ 0,x]$ such that
begin{align} cosleft(dfrac{x}{k}right)-cosleft(0right) =-dfrac{1}{k}sinleft(dfrac{c}{k}right),(x-0), end{align}
which implies begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right| &=left|sum^{infty}_{k=1}dfrac{x}{k}sinleft(dfrac{c}{k}right)right| leq sum^{infty}_{k=1}dfrac{left|xright|}{k}left|sinleft(dfrac{c}{k}right)right|leq sum^{infty}_{k=1}dfrac{left|xright|}{k}dfrac{left|cright|}{k}\&leq sum^{infty}_{k=1}left(dfrac{left|xright|}{k}right)^2leq sum^{infty}_{k=1}dfrac{1}{k^2}=1+ sum^{infty}_{k=2}dfrac{1}{k^2}\&leq 1+ sum^{infty}_{k=2}dfrac{1}{k(k-1)}\&= 1+ limlimits_{ntoinfty}sum^{n}_{k=2}left(dfrac{1}{k-1}-dfrac{1}{k}right)\&=1+ limlimits_{ntoinfty}left(1-dfrac{1}{n}right)\&=2, end{align}



METHOD 2



Let $xin [0,1]$ be fixed, then
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}dfrac{1}{k}left[-kcosleft(dfrac{x}{k}right)right]^{1}_{0}=sum^{infty}_{k=1}dfrac{1}{k}int^{1}_{0}sinleft(dfrac{x}{k}right)dx \&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx end{align}
The series $sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)$ converges uniformly on $[0,1]$, by Weierstrass-M test, since
begin{align} left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k^2}. end{align}
Hence,
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx=int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx, end{align}
and
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&=left|int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dxright|leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right)right|dx \&leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k^2}dx \&leq 2 end{align}










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$endgroup$












  • $begingroup$
    Nice work! (+1) Do you know if the series $$sum_{ngeq1}1-cos(1/k)$$ Has a closed form?
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:14










  • $begingroup$
    Hmm... I don't know but do you have an idea its closed form?
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:16












  • $begingroup$
    I do not. It looks a lot like a Fourier series though. Special functions may help...
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:21












  • $begingroup$
    @clathratus: Kindly see Robert Israel's answer for a closed form.
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:35










  • $begingroup$
    Hi @Mike ! I would like to learn how to evaluate to solve some problems like the one you posted, but I've never seen such problems in my Analysis books. Could you please share how you learned to solve them and if there is any resource with many problems like this, either teaching you how to do them or just many practice problems?
    $endgroup$
    – Ovi
    Jan 1 at 0:36














7












7








7


5



$begingroup$


I've found two ways of proving that
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&leq 2 end{align}
Are there any other ways out there, for proving this?



METHOD 1



Let $kin Bbb{N}$, then



begin{align} f:[ 0&,1]longrightarrow Bbb{R}\&x mapsto
cosleft(dfrac{x}{k}right) end{align}

is continuous. Then, by Mean Value Theorem, there exists $cin [ 0,x]$ such that
begin{align} cosleft(dfrac{x}{k}right)-cosleft(0right) =-dfrac{1}{k}sinleft(dfrac{c}{k}right),(x-0), end{align}
which implies begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right| &=left|sum^{infty}_{k=1}dfrac{x}{k}sinleft(dfrac{c}{k}right)right| leq sum^{infty}_{k=1}dfrac{left|xright|}{k}left|sinleft(dfrac{c}{k}right)right|leq sum^{infty}_{k=1}dfrac{left|xright|}{k}dfrac{left|cright|}{k}\&leq sum^{infty}_{k=1}left(dfrac{left|xright|}{k}right)^2leq sum^{infty}_{k=1}dfrac{1}{k^2}=1+ sum^{infty}_{k=2}dfrac{1}{k^2}\&leq 1+ sum^{infty}_{k=2}dfrac{1}{k(k-1)}\&= 1+ limlimits_{ntoinfty}sum^{n}_{k=2}left(dfrac{1}{k-1}-dfrac{1}{k}right)\&=1+ limlimits_{ntoinfty}left(1-dfrac{1}{n}right)\&=2, end{align}



METHOD 2



Let $xin [0,1]$ be fixed, then
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}dfrac{1}{k}left[-kcosleft(dfrac{x}{k}right)right]^{1}_{0}=sum^{infty}_{k=1}dfrac{1}{k}int^{1}_{0}sinleft(dfrac{x}{k}right)dx \&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx end{align}
The series $sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)$ converges uniformly on $[0,1]$, by Weierstrass-M test, since
begin{align} left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k^2}. end{align}
Hence,
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx=int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx, end{align}
and
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&=left|int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dxright|leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right)right|dx \&leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k^2}dx \&leq 2 end{align}










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$endgroup$




I've found two ways of proving that
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&leq 2 end{align}
Are there any other ways out there, for proving this?



METHOD 1



Let $kin Bbb{N}$, then



begin{align} f:[ 0&,1]longrightarrow Bbb{R}\&x mapsto
cosleft(dfrac{x}{k}right) end{align}

is continuous. Then, by Mean Value Theorem, there exists $cin [ 0,x]$ such that
begin{align} cosleft(dfrac{x}{k}right)-cosleft(0right) =-dfrac{1}{k}sinleft(dfrac{c}{k}right),(x-0), end{align}
which implies begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right| &=left|sum^{infty}_{k=1}dfrac{x}{k}sinleft(dfrac{c}{k}right)right| leq sum^{infty}_{k=1}dfrac{left|xright|}{k}left|sinleft(dfrac{c}{k}right)right|leq sum^{infty}_{k=1}dfrac{left|xright|}{k}dfrac{left|cright|}{k}\&leq sum^{infty}_{k=1}left(dfrac{left|xright|}{k}right)^2leq sum^{infty}_{k=1}dfrac{1}{k^2}=1+ sum^{infty}_{k=2}dfrac{1}{k^2}\&leq 1+ sum^{infty}_{k=2}dfrac{1}{k(k-1)}\&= 1+ limlimits_{ntoinfty}sum^{n}_{k=2}left(dfrac{1}{k-1}-dfrac{1}{k}right)\&=1+ limlimits_{ntoinfty}left(1-dfrac{1}{n}right)\&=2, end{align}



METHOD 2



Let $xin [0,1]$ be fixed, then
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}dfrac{1}{k}left[-kcosleft(dfrac{x}{k}right)right]^{1}_{0}=sum^{infty}_{k=1}dfrac{1}{k}int^{1}_{0}sinleft(dfrac{x}{k}right)dx \&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx end{align}
The series $sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)$ converges uniformly on $[0,1]$, by Weierstrass-M test, since
begin{align} left|sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right) right|leq sum^{infty}_{k=1}dfrac{1}{k^2}. end{align}
Hence,
begin{align} sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]&=sum^{infty}_{k=1}int^{1}_{0}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx=int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dx, end{align}
and
begin{align} left|sum^{infty}_{k=1}left[1-cosleft(dfrac{1}{k}right)right]right|&=left|int^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}sinleft(dfrac{x}{k}right)dxright|leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k}left|sinleft(dfrac{x}{k}right)right|dx \&leqint^{1}_{0}sum^{infty}_{k=1}dfrac{1}{k^2}dx \&leq 2 end{align}







real-analysis analysis convergence uniform-convergence






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edited Dec 31 '18 at 20:10







Omojola Micheal

















asked Dec 30 '18 at 23:07









Omojola MichealOmojola Micheal

1,986424




1,986424












  • $begingroup$
    Nice work! (+1) Do you know if the series $$sum_{ngeq1}1-cos(1/k)$$ Has a closed form?
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:14










  • $begingroup$
    Hmm... I don't know but do you have an idea its closed form?
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:16












  • $begingroup$
    I do not. It looks a lot like a Fourier series though. Special functions may help...
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:21












  • $begingroup$
    @clathratus: Kindly see Robert Israel's answer for a closed form.
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:35










  • $begingroup$
    Hi @Mike ! I would like to learn how to evaluate to solve some problems like the one you posted, but I've never seen such problems in my Analysis books. Could you please share how you learned to solve them and if there is any resource with many problems like this, either teaching you how to do them or just many practice problems?
    $endgroup$
    – Ovi
    Jan 1 at 0:36


















  • $begingroup$
    Nice work! (+1) Do you know if the series $$sum_{ngeq1}1-cos(1/k)$$ Has a closed form?
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:14










  • $begingroup$
    Hmm... I don't know but do you have an idea its closed form?
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:16












  • $begingroup$
    I do not. It looks a lot like a Fourier series though. Special functions may help...
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:21












  • $begingroup$
    @clathratus: Kindly see Robert Israel's answer for a closed form.
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:35










  • $begingroup$
    Hi @Mike ! I would like to learn how to evaluate to solve some problems like the one you posted, but I've never seen such problems in my Analysis books. Could you please share how you learned to solve them and if there is any resource with many problems like this, either teaching you how to do them or just many practice problems?
    $endgroup$
    – Ovi
    Jan 1 at 0:36
















$begingroup$
Nice work! (+1) Do you know if the series $$sum_{ngeq1}1-cos(1/k)$$ Has a closed form?
$endgroup$
– clathratus
Dec 30 '18 at 23:14




$begingroup$
Nice work! (+1) Do you know if the series $$sum_{ngeq1}1-cos(1/k)$$ Has a closed form?
$endgroup$
– clathratus
Dec 30 '18 at 23:14












$begingroup$
Hmm... I don't know but do you have an idea its closed form?
$endgroup$
– Omojola Micheal
Dec 30 '18 at 23:16






$begingroup$
Hmm... I don't know but do you have an idea its closed form?
$endgroup$
– Omojola Micheal
Dec 30 '18 at 23:16














$begingroup$
I do not. It looks a lot like a Fourier series though. Special functions may help...
$endgroup$
– clathratus
Dec 30 '18 at 23:21






$begingroup$
I do not. It looks a lot like a Fourier series though. Special functions may help...
$endgroup$
– clathratus
Dec 30 '18 at 23:21














$begingroup$
@clathratus: Kindly see Robert Israel's answer for a closed form.
$endgroup$
– Omojola Micheal
Dec 30 '18 at 23:35




$begingroup$
@clathratus: Kindly see Robert Israel's answer for a closed form.
$endgroup$
– Omojola Micheal
Dec 30 '18 at 23:35












$begingroup$
Hi @Mike ! I would like to learn how to evaluate to solve some problems like the one you posted, but I've never seen such problems in my Analysis books. Could you please share how you learned to solve them and if there is any resource with many problems like this, either teaching you how to do them or just many practice problems?
$endgroup$
– Ovi
Jan 1 at 0:36




$begingroup$
Hi @Mike ! I would like to learn how to evaluate to solve some problems like the one you posted, but I've never seen such problems in my Analysis books. Could you please share how you learned to solve them and if there is any resource with many problems like this, either teaching you how to do them or just many practice problems?
$endgroup$
– Ovi
Jan 1 at 0:36










4 Answers
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Using the Maclaurin series for $cos$,
your sum is $$S = sum_{k=1}^infty sum_{j=1}^infty frac{(-1)^{j+1}}{(2j)!; k^{2j}}$$
This converges absolutely, and
$$ |S| le sum_{j=1}^infty sum_{k=1}^infty frac{1}{(2j)!; k^{2j}} = sum_{j=1}^infty frac{zeta(2j)}{(2j)!} le sum_{j=1}^infty frac{zeta(2)}{(2j)!} =frac{(cosh(1)-1) pi^2}{6} < 2$$
(in fact $< 0.9$).






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  • $begingroup$
    I love this method!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:27



















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Note that
$$1 - cos(x) = cos(0) - cos(x) = 2 sin^2left( frac{x}2 right) $$
by the sum to product identities for all $x$. Therefore,



$$ sum_{k=1}^{infty} left(1- cosleft( frac{1}k right) right) = 2 sum_{k=1}^{infty} sin^2left( frac{1}{2k} right). $$
Now using the inequality $sin(x) le x$ for all positive $x$, we get
$$ 2sum_{k=1}^{infty} sin^2left( frac{1}{2k} right) le frac{1}{2}sum_{k=1}^{infty} frac{1}{k^2} approx 0.822.$$






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  • $begingroup$
    You should say $|sin x| le |x|$ for all $x.$ (Minor quibble.)
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 22:52










  • $begingroup$
    @DanielWainfleet Thanks I made it slightly better
    $endgroup$
    – Sandeep Silwal
    Jan 1 at 0:25



















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Another method which gives a better bound than $2$ is the following: the inequality $1-cos, x leq frac {x^{2}} 2$ holds for all real $x$ and $sum frac 1 {k^{2}} =frac {pi^{2}} 6$. Use the fact that $frac {pi^{2}} {12}<2$. [ $1-cos, x - frac {x^{2}} 2$ vanishes at $0$ and its derivative is $sin, x -x <0$. This gives the inequality above]. Note that LHS $<1$ in fact!






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  • $begingroup$
    I did not downvote, but perhaps the down vote is due to the similarity to the second method above? Although this is perhaps cleaner and more clear, too.
    $endgroup$
    – Clayton
    Dec 30 '18 at 23:21










  • $begingroup$
    @Clayton right... Thanks. It's not my day :(
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:26










  • $begingroup$
    I like this approach!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:29



















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$begingroup$

Using the Taylor series of the $cos$ function, your sum is equal to the much faster converging $sum_{nge 1} (-1)^{n+1}frac{zeta(2n)}{(2n)!}$. Summing the first two terms, $pi^2/12-pi^4/2160$, gives a value of $approx 0.7773$ with an error of $epsilon_1approx1.388times 10^{-3}$. By contrast, summing two terms of your original sum will give you $approx 0.5821$, this time with a larger error of $epsilon_2 approx .1942$.



To answer your question, we can use a theorem about alternating series, which states that for $S_n=sum_{k=1}^n (-1)^{k+1} a_k$ and $S=lim_{ntoinfty}S_n$, we have that
$$|S-S_n|< a_{n+1}$$
Using the same two terms as above, we can see that $S<pi^6/680400 - pi^4/2160 + pi^2/12approx 0.7788$. Thus,



$$sum_{k=1}^infty 1-cos(1/k) < 0.78 <2$$






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  • $begingroup$
    (+1) for that. Looks good!
    $endgroup$
    – Omojola Micheal
    Dec 31 '18 at 1:17











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4 Answers
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4 Answers
4






active

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8












$begingroup$

Using the Maclaurin series for $cos$,
your sum is $$S = sum_{k=1}^infty sum_{j=1}^infty frac{(-1)^{j+1}}{(2j)!; k^{2j}}$$
This converges absolutely, and
$$ |S| le sum_{j=1}^infty sum_{k=1}^infty frac{1}{(2j)!; k^{2j}} = sum_{j=1}^infty frac{zeta(2j)}{(2j)!} le sum_{j=1}^infty frac{zeta(2)}{(2j)!} =frac{(cosh(1)-1) pi^2}{6} < 2$$
(in fact $< 0.9$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I love this method!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:27
















8












$begingroup$

Using the Maclaurin series for $cos$,
your sum is $$S = sum_{k=1}^infty sum_{j=1}^infty frac{(-1)^{j+1}}{(2j)!; k^{2j}}$$
This converges absolutely, and
$$ |S| le sum_{j=1}^infty sum_{k=1}^infty frac{1}{(2j)!; k^{2j}} = sum_{j=1}^infty frac{zeta(2j)}{(2j)!} le sum_{j=1}^infty frac{zeta(2)}{(2j)!} =frac{(cosh(1)-1) pi^2}{6} < 2$$
(in fact $< 0.9$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I love this method!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:27














8












8








8





$begingroup$

Using the Maclaurin series for $cos$,
your sum is $$S = sum_{k=1}^infty sum_{j=1}^infty frac{(-1)^{j+1}}{(2j)!; k^{2j}}$$
This converges absolutely, and
$$ |S| le sum_{j=1}^infty sum_{k=1}^infty frac{1}{(2j)!; k^{2j}} = sum_{j=1}^infty frac{zeta(2j)}{(2j)!} le sum_{j=1}^infty frac{zeta(2)}{(2j)!} =frac{(cosh(1)-1) pi^2}{6} < 2$$
(in fact $< 0.9$).






share|cite|improve this answer









$endgroup$



Using the Maclaurin series for $cos$,
your sum is $$S = sum_{k=1}^infty sum_{j=1}^infty frac{(-1)^{j+1}}{(2j)!; k^{2j}}$$
This converges absolutely, and
$$ |S| le sum_{j=1}^infty sum_{k=1}^infty frac{1}{(2j)!; k^{2j}} = sum_{j=1}^infty frac{zeta(2j)}{(2j)!} le sum_{j=1}^infty frac{zeta(2)}{(2j)!} =frac{(cosh(1)-1) pi^2}{6} < 2$$
(in fact $< 0.9$).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 23:26









Robert IsraelRobert Israel

329k23217470




329k23217470












  • $begingroup$
    I love this method!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:27


















  • $begingroup$
    I love this method!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:27
















$begingroup$
I love this method!
$endgroup$
– Omojola Micheal
Dec 30 '18 at 23:27




$begingroup$
I love this method!
$endgroup$
– Omojola Micheal
Dec 30 '18 at 23:27











8












$begingroup$

Note that
$$1 - cos(x) = cos(0) - cos(x) = 2 sin^2left( frac{x}2 right) $$
by the sum to product identities for all $x$. Therefore,



$$ sum_{k=1}^{infty} left(1- cosleft( frac{1}k right) right) = 2 sum_{k=1}^{infty} sin^2left( frac{1}{2k} right). $$
Now using the inequality $sin(x) le x$ for all positive $x$, we get
$$ 2sum_{k=1}^{infty} sin^2left( frac{1}{2k} right) le frac{1}{2}sum_{k=1}^{infty} frac{1}{k^2} approx 0.822.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You should say $|sin x| le |x|$ for all $x.$ (Minor quibble.)
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 22:52










  • $begingroup$
    @DanielWainfleet Thanks I made it slightly better
    $endgroup$
    – Sandeep Silwal
    Jan 1 at 0:25
















8












$begingroup$

Note that
$$1 - cos(x) = cos(0) - cos(x) = 2 sin^2left( frac{x}2 right) $$
by the sum to product identities for all $x$. Therefore,



$$ sum_{k=1}^{infty} left(1- cosleft( frac{1}k right) right) = 2 sum_{k=1}^{infty} sin^2left( frac{1}{2k} right). $$
Now using the inequality $sin(x) le x$ for all positive $x$, we get
$$ 2sum_{k=1}^{infty} sin^2left( frac{1}{2k} right) le frac{1}{2}sum_{k=1}^{infty} frac{1}{k^2} approx 0.822.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You should say $|sin x| le |x|$ for all $x.$ (Minor quibble.)
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 22:52










  • $begingroup$
    @DanielWainfleet Thanks I made it slightly better
    $endgroup$
    – Sandeep Silwal
    Jan 1 at 0:25














8












8








8





$begingroup$

Note that
$$1 - cos(x) = cos(0) - cos(x) = 2 sin^2left( frac{x}2 right) $$
by the sum to product identities for all $x$. Therefore,



$$ sum_{k=1}^{infty} left(1- cosleft( frac{1}k right) right) = 2 sum_{k=1}^{infty} sin^2left( frac{1}{2k} right). $$
Now using the inequality $sin(x) le x$ for all positive $x$, we get
$$ 2sum_{k=1}^{infty} sin^2left( frac{1}{2k} right) le frac{1}{2}sum_{k=1}^{infty} frac{1}{k^2} approx 0.822.$$






share|cite|improve this answer











$endgroup$



Note that
$$1 - cos(x) = cos(0) - cos(x) = 2 sin^2left( frac{x}2 right) $$
by the sum to product identities for all $x$. Therefore,



$$ sum_{k=1}^{infty} left(1- cosleft( frac{1}k right) right) = 2 sum_{k=1}^{infty} sin^2left( frac{1}{2k} right). $$
Now using the inequality $sin(x) le x$ for all positive $x$, we get
$$ 2sum_{k=1}^{infty} sin^2left( frac{1}{2k} right) le frac{1}{2}sum_{k=1}^{infty} frac{1}{k^2} approx 0.822.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 1 at 0:24

























answered Dec 31 '18 at 4:55









Sandeep SilwalSandeep Silwal

5,88811237




5,88811237












  • $begingroup$
    You should say $|sin x| le |x|$ for all $x.$ (Minor quibble.)
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 22:52










  • $begingroup$
    @DanielWainfleet Thanks I made it slightly better
    $endgroup$
    – Sandeep Silwal
    Jan 1 at 0:25


















  • $begingroup$
    You should say $|sin x| le |x|$ for all $x.$ (Minor quibble.)
    $endgroup$
    – DanielWainfleet
    Dec 31 '18 at 22:52










  • $begingroup$
    @DanielWainfleet Thanks I made it slightly better
    $endgroup$
    – Sandeep Silwal
    Jan 1 at 0:25
















$begingroup$
You should say $|sin x| le |x|$ for all $x.$ (Minor quibble.)
$endgroup$
– DanielWainfleet
Dec 31 '18 at 22:52




$begingroup$
You should say $|sin x| le |x|$ for all $x.$ (Minor quibble.)
$endgroup$
– DanielWainfleet
Dec 31 '18 at 22:52












$begingroup$
@DanielWainfleet Thanks I made it slightly better
$endgroup$
– Sandeep Silwal
Jan 1 at 0:25




$begingroup$
@DanielWainfleet Thanks I made it slightly better
$endgroup$
– Sandeep Silwal
Jan 1 at 0:25











4












$begingroup$

Another method which gives a better bound than $2$ is the following: the inequality $1-cos, x leq frac {x^{2}} 2$ holds for all real $x$ and $sum frac 1 {k^{2}} =frac {pi^{2}} 6$. Use the fact that $frac {pi^{2}} {12}<2$. [ $1-cos, x - frac {x^{2}} 2$ vanishes at $0$ and its derivative is $sin, x -x <0$. This gives the inequality above]. Note that LHS $<1$ in fact!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I did not downvote, but perhaps the down vote is due to the similarity to the second method above? Although this is perhaps cleaner and more clear, too.
    $endgroup$
    – Clayton
    Dec 30 '18 at 23:21










  • $begingroup$
    @Clayton right... Thanks. It's not my day :(
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:26










  • $begingroup$
    I like this approach!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:29
















4












$begingroup$

Another method which gives a better bound than $2$ is the following: the inequality $1-cos, x leq frac {x^{2}} 2$ holds for all real $x$ and $sum frac 1 {k^{2}} =frac {pi^{2}} 6$. Use the fact that $frac {pi^{2}} {12}<2$. [ $1-cos, x - frac {x^{2}} 2$ vanishes at $0$ and its derivative is $sin, x -x <0$. This gives the inequality above]. Note that LHS $<1$ in fact!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I did not downvote, but perhaps the down vote is due to the similarity to the second method above? Although this is perhaps cleaner and more clear, too.
    $endgroup$
    – Clayton
    Dec 30 '18 at 23:21










  • $begingroup$
    @Clayton right... Thanks. It's not my day :(
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:26










  • $begingroup$
    I like this approach!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:29














4












4








4





$begingroup$

Another method which gives a better bound than $2$ is the following: the inequality $1-cos, x leq frac {x^{2}} 2$ holds for all real $x$ and $sum frac 1 {k^{2}} =frac {pi^{2}} 6$. Use the fact that $frac {pi^{2}} {12}<2$. [ $1-cos, x - frac {x^{2}} 2$ vanishes at $0$ and its derivative is $sin, x -x <0$. This gives the inequality above]. Note that LHS $<1$ in fact!






share|cite|improve this answer











$endgroup$



Another method which gives a better bound than $2$ is the following: the inequality $1-cos, x leq frac {x^{2}} 2$ holds for all real $x$ and $sum frac 1 {k^{2}} =frac {pi^{2}} 6$. Use the fact that $frac {pi^{2}} {12}<2$. [ $1-cos, x - frac {x^{2}} 2$ vanishes at $0$ and its derivative is $sin, x -x <0$. This gives the inequality above]. Note that LHS $<1$ in fact!







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 23:24

























answered Dec 30 '18 at 23:17









Kavi Rama MurthyKavi Rama Murthy

69.9k53170




69.9k53170












  • $begingroup$
    I did not downvote, but perhaps the down vote is due to the similarity to the second method above? Although this is perhaps cleaner and more clear, too.
    $endgroup$
    – Clayton
    Dec 30 '18 at 23:21










  • $begingroup$
    @Clayton right... Thanks. It's not my day :(
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:26










  • $begingroup$
    I like this approach!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:29


















  • $begingroup$
    I did not downvote, but perhaps the down vote is due to the similarity to the second method above? Although this is perhaps cleaner and more clear, too.
    $endgroup$
    – Clayton
    Dec 30 '18 at 23:21










  • $begingroup$
    @Clayton right... Thanks. It's not my day :(
    $endgroup$
    – clathratus
    Dec 30 '18 at 23:26










  • $begingroup$
    I like this approach!
    $endgroup$
    – Omojola Micheal
    Dec 30 '18 at 23:29
















$begingroup$
I did not downvote, but perhaps the down vote is due to the similarity to the second method above? Although this is perhaps cleaner and more clear, too.
$endgroup$
– Clayton
Dec 30 '18 at 23:21




$begingroup$
I did not downvote, but perhaps the down vote is due to the similarity to the second method above? Although this is perhaps cleaner and more clear, too.
$endgroup$
– Clayton
Dec 30 '18 at 23:21












$begingroup$
@Clayton right... Thanks. It's not my day :(
$endgroup$
– clathratus
Dec 30 '18 at 23:26




$begingroup$
@Clayton right... Thanks. It's not my day :(
$endgroup$
– clathratus
Dec 30 '18 at 23:26












$begingroup$
I like this approach!
$endgroup$
– Omojola Micheal
Dec 30 '18 at 23:29




$begingroup$
I like this approach!
$endgroup$
– Omojola Micheal
Dec 30 '18 at 23:29











4












$begingroup$

Using the Taylor series of the $cos$ function, your sum is equal to the much faster converging $sum_{nge 1} (-1)^{n+1}frac{zeta(2n)}{(2n)!}$. Summing the first two terms, $pi^2/12-pi^4/2160$, gives a value of $approx 0.7773$ with an error of $epsilon_1approx1.388times 10^{-3}$. By contrast, summing two terms of your original sum will give you $approx 0.5821$, this time with a larger error of $epsilon_2 approx .1942$.



To answer your question, we can use a theorem about alternating series, which states that for $S_n=sum_{k=1}^n (-1)^{k+1} a_k$ and $S=lim_{ntoinfty}S_n$, we have that
$$|S-S_n|< a_{n+1}$$
Using the same two terms as above, we can see that $S<pi^6/680400 - pi^4/2160 + pi^2/12approx 0.7788$. Thus,



$$sum_{k=1}^infty 1-cos(1/k) < 0.78 <2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) for that. Looks good!
    $endgroup$
    – Omojola Micheal
    Dec 31 '18 at 1:17
















4












$begingroup$

Using the Taylor series of the $cos$ function, your sum is equal to the much faster converging $sum_{nge 1} (-1)^{n+1}frac{zeta(2n)}{(2n)!}$. Summing the first two terms, $pi^2/12-pi^4/2160$, gives a value of $approx 0.7773$ with an error of $epsilon_1approx1.388times 10^{-3}$. By contrast, summing two terms of your original sum will give you $approx 0.5821$, this time with a larger error of $epsilon_2 approx .1942$.



To answer your question, we can use a theorem about alternating series, which states that for $S_n=sum_{k=1}^n (-1)^{k+1} a_k$ and $S=lim_{ntoinfty}S_n$, we have that
$$|S-S_n|< a_{n+1}$$
Using the same two terms as above, we can see that $S<pi^6/680400 - pi^4/2160 + pi^2/12approx 0.7788$. Thus,



$$sum_{k=1}^infty 1-cos(1/k) < 0.78 <2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) for that. Looks good!
    $endgroup$
    – Omojola Micheal
    Dec 31 '18 at 1:17














4












4








4





$begingroup$

Using the Taylor series of the $cos$ function, your sum is equal to the much faster converging $sum_{nge 1} (-1)^{n+1}frac{zeta(2n)}{(2n)!}$. Summing the first two terms, $pi^2/12-pi^4/2160$, gives a value of $approx 0.7773$ with an error of $epsilon_1approx1.388times 10^{-3}$. By contrast, summing two terms of your original sum will give you $approx 0.5821$, this time with a larger error of $epsilon_2 approx .1942$.



To answer your question, we can use a theorem about alternating series, which states that for $S_n=sum_{k=1}^n (-1)^{k+1} a_k$ and $S=lim_{ntoinfty}S_n$, we have that
$$|S-S_n|< a_{n+1}$$
Using the same two terms as above, we can see that $S<pi^6/680400 - pi^4/2160 + pi^2/12approx 0.7788$. Thus,



$$sum_{k=1}^infty 1-cos(1/k) < 0.78 <2$$






share|cite|improve this answer











$endgroup$



Using the Taylor series of the $cos$ function, your sum is equal to the much faster converging $sum_{nge 1} (-1)^{n+1}frac{zeta(2n)}{(2n)!}$. Summing the first two terms, $pi^2/12-pi^4/2160$, gives a value of $approx 0.7773$ with an error of $epsilon_1approx1.388times 10^{-3}$. By contrast, summing two terms of your original sum will give you $approx 0.5821$, this time with a larger error of $epsilon_2 approx .1942$.



To answer your question, we can use a theorem about alternating series, which states that for $S_n=sum_{k=1}^n (-1)^{k+1} a_k$ and $S=lim_{ntoinfty}S_n$, we have that
$$|S-S_n|< a_{n+1}$$
Using the same two terms as above, we can see that $S<pi^6/680400 - pi^4/2160 + pi^2/12approx 0.7788$. Thus,



$$sum_{k=1}^infty 1-cos(1/k) < 0.78 <2$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 20:55

























answered Dec 31 '18 at 1:15









ZacharyZachary

2,3751214




2,3751214












  • $begingroup$
    (+1) for that. Looks good!
    $endgroup$
    – Omojola Micheal
    Dec 31 '18 at 1:17


















  • $begingroup$
    (+1) for that. Looks good!
    $endgroup$
    – Omojola Micheal
    Dec 31 '18 at 1:17
















$begingroup$
(+1) for that. Looks good!
$endgroup$
– Omojola Micheal
Dec 31 '18 at 1:17




$begingroup$
(+1) for that. Looks good!
$endgroup$
– Omojola Micheal
Dec 31 '18 at 1:17


















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