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Showing convergence is easy, since for $x\geq 0$ we have $\sin x\leq x$ so:

$$0\leq\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{1}{n}\right)\leq \sum_{n=1}^{\infty}\frac{1}{n^2}$$

I was wondering if it might have a closed form? Finding one is beyond me, but I am sure that the people on here will have ideas.

Mithlesh Upadhyay
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Little miss sunshine
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    try Taylor expansion, you will get $\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \dfrac{(-1)^{(k-1)}}{(2k-1)!}\dfrac{1}{n^{2k}}=\sum_{k=1}^{\infty}\dfrac{(-1)^{(k-1)}}{(2k-1)!}\zeta(2k)$ – Yimin Mar 08 '13 at 01:15
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    My idea is that there is no reason to expect a closed form. – Will Jagy Mar 08 '13 at 01:54

1 Answers1

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The sum is approximately $1.47282823195618529629494738$. Plugging this into the Inverse Symbolic Calculator fails to find a match. Most likely, this has no closed form.

robjohn
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  • The 'algebraized' version $\sum\dfrac1n\sin\left(\dfrac\pi n\right)$ also seems to have no closed form - I get a value of roughly 1.6425614523 from the zeta method, and the ISC fails to find anything useful here either. – Steven Stadnicki Mar 08 '13 at 03:06
  • @StevenStadnicki: What do you mean 'algebraized'? How did you compute that value? – robjohn Mar 08 '13 at 03:55
  • 'algebraized' just in that all of the individual terms are algebraic (as opposed to the OP's original sum, which has all the terms transcendental); this doesn't mean anything particular, of course, but I was curious. I computed it via the same Taylor expansion as mentioned in Yimin's comment to the original post - it's equal to the sum $\sum_{k=0}^\infty\dfrac{(-1)^k\pi^{2k+1}}{(2k+1)!}\zeta(2k+2)$, and this converges quickly enough to evaluate to many places with Alpha. – Steven Stadnicki Mar 08 '13 at 17:01
  • @StevenStadnicki: Ah, I see the difference. You have $\pi^{2k+1}$ in yours, but Yimin doesn't. – robjohn Mar 08 '13 at 18:29
  • Yimin's is the formula for $\sum\dfrac1n\sin(\dfrac1n)$; mine is the formula for $\sum\dfrac1n\sin(\dfrac\pi n)$ (as I said, the 'algebraized' version - I'm not expecting it to be the same value, of course, but was curious if it had a closed form either). Both are correct for their respective sums. – Steven Stadnicki Mar 08 '13 at 18:33