TITLE: Evaluating a sum to infinity
QUESTION [7 upvotes]: 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.

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