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I'm looking for a way that allows me to work out the following sum:

$$\sum\limits_{k=1}^{\infty} \sin^2\left(\frac{1}{k}\right)$$

Any hint/suggestion is welcome. Thanks.

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It almost certainly doesn't have a recognizable closed form. Would you settle for an argument that it converges? –  Qiaochu Yuan Jun 8 '12 at 19:59
2  
@Qiaochu Yuan: i want to find its exact limit. It's easy to prove its convergence. –  Chris's sis Jun 8 '12 at 20:01
    
The only thing can tink of off the top of my head is that sin(x) is approximately x for small x. So, for large enough k, your terms are approximately $1/k^2$. Now, limit comparison should finish it off. That will get you covergence, but not the sum. Disregard, I just saw your response to Qiaochu. –  Chris Leary Jun 8 '12 at 20:02
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Almost surely this doesn't have a "nice" value as its limit. –  user17762 Jun 8 '12 at 20:06
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It's value is 1.326324405266..., among the 54 million known constants i checked, it wasnt there. –  user1708 Jun 8 '12 at 20:16

1 Answer 1

up vote 9 down vote accepted

It may be too much to ask for a closed form. We find an equivalent series that converges very fast. We have $$\begin{eqnarray*} \sum_{k=1}^\infty \sin^2\frac{1}{k} &=& \sum_{k=1}^\infty \frac{1}{2}\left(1-\cos \frac{2}{k}\right) \\ &=& \frac{1}{2} \sum_{k=1}^\infty \sum_{j=1}^\infty \frac{(-1)^{j+1}}{(2j)!} \left(\frac{2}{k}\right)^{2j} \\ &=& \frac{1}{2} \sum_{j=1}^\infty \frac{(-1)^{j+1} 2^{2j}}{(2j)!} \zeta(2j) \\ &=& \frac{1}{4} \sum_{j=1}^\infty \frac{(4\pi)^{2j}}{[(2j)!]^2} B_{2j} \end{eqnarray*}$$ where $\zeta(2j)$ is the zeta function and $B_{2j}$ are the Bernoulli numbers. Interchanging the sums is allowed by Fubini's theorem. The ratio of successive terms goes like $1/j^2$ for $j$ large.

Below we give the partial sums to $25$ digits. $$\begin{array}{ll} N & \frac{1}{4} \sum_{j=1}^N \frac{(4\pi)^{2j}}{[(2j)!]^2} B_{2j}\\\hline 1 & 1.644934066848226436472415\cdots \\ 2 & 1.284159655611180372633747\cdots \\ 3 & 1.329374902810489223287726\cdots \\ 4 & 1.326187355647956066654778\cdots \\ 5 & 1.326328589450443236755002\cdots \\ 6 & 1.326324312838454339066804\cdots \\ 7 & 1.326324406812557661734373\cdots \\ 8 & 1.326324405246394595313185\cdots \\ 9 & 1.326324405266867080420232\cdots \\ 10 & 1.326324405266651581194045\cdots \\ 11 & 1.326324405266653446986876\cdots \\ 12 & 1.326324405266653433466641\cdots \\ 13 & 1.326324405266653433549842\cdots \\ 14 & 1.326324405266653433549402\cdots \\ 15 & 1.326324405266653433549404\cdots \end{array}$$

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nice work there! –  Chris's sis Jun 8 '12 at 21:32
    
@Chris: Glad to help. Cheers! –  user26872 Jun 8 '12 at 21:34
    
yeah. Thanks for your great approach. –  Chris's sis Jun 8 '12 at 21:35
    
very glad to see that there is a way to tackle all limits :) –  Chris's sis Jun 8 '12 at 21:47

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