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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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1  
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. –  anon Jul 16 '11 at 10:25
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2 Answers

up vote 3 down vote accepted

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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I also doubt there's a closed form, but as far as you intended to imply that this is related to the harmonic numbers not having a closed form, I'd disagree. $\log(n-1)+\sum_{k=1}^\infty(k(1-n)^k)^{-1}$ also gives values close to the harmonic numbers, yet has the closed form $\log n$. –  joriki Jul 16 '11 at 9:49
    
@joriki: I would agree. I was pointing to the harmonic numbers as an example of something not having a closed form. –  Henry Jul 16 '11 at 9:54
    
I consider "harmonic numbers" and "polygamma functions" as closed forms, but even with that expanded notion, there seems to be no applicable "known function" for that sine summation... it looks terribly unnatural to me in any event. –  J. M. Jul 16 '11 at 12:12
    
Correct. This sum is unnatural. It has no known "closed form". And (until the OP reveals his secret) no known applications either! –  GEdgar Jul 16 '11 at 13:13
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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.

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