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I'm trying to find closed form for

$\sum_{k=1}^{n}\sin\frac{1}{k}$

I typed it in Mathematica 6.0 and WolframAlpha, but no result what i expected.

Any hints will be appreciated, thank you.

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    I deleted my answer because I misread it as an infinite series. In any case, if you label your sum $S_n$, then you can use the Taylor expansion of $\sin x$ to write the difference between harmonic numbers and your sum with the Riemann zeta function: $ (H_n - S_n) \to \frac{\zeta(3)}{3!} - \frac{\zeta(5)}{5!} + \frac{\zeta(7)}{7!} - \cdots$ I'm not sure what the error term is, but it's probably pretty small.2011-07-16

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I doubt you will find a closed form.

Your expression will give a value slightly less than than the harmonic numbers $\sum_{k=1}^{n}\frac{1}{k}$ which do not have a closed form and as $k$ increases the difference will increase towards $0.191899\ldots$, and a value slightly more than $\log n$ and as $k$ increases the difference will fall towards $0.385316\ldots$.

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    Correct. This sum is unnatural. It has no known "closed form". And (until the OP reveals his secret) no known applications either!2011-07-16
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The sum can be expanded in the asymptotic series, several first members being $ \sum_{k=1}^n\sin\frac{1}{k}= \log n+a+\frac1{2n}-\frac1{12n^3}+O\left(\frac1{n^3}\right), $ where $ a=\gamma+\sum_{k=1}^\infty (-1)^k\frac{\zeta(2k+1)}{(2k+1)!} $ and $\gamma$ is the Euler constant. The value of $a$ is $0.38...$ as written by Henry.