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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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    The inverse symbolic calculator agrees: [link](http://oldweb.cecm.sfu.ca/cgi-bin/isc/lookup?number=sum%28sin%281%2Fk%29^2%2Ck%3D1..infinity%29&lookup_type=simple)2012-06-08

1 Answers 1

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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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    very glad to see that there is a way to tackle all limits :)2012-06-08