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Does the series $\sum_{k=1}^\infty\frac{\sin(1/k)}{k}$ converge?

By Taylor expanding, I see that this can be rewritten as

$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+1}}{k^{2n}(2n-1)!}$,

but that seems to be making it messier than it needs to be.

Can we use the limit comparison test here?

Any help appreciated!

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    $\begingroup$ How about using $\sin(n)<n$? $\endgroup$ – Simply Beautiful Art Feb 25 '17 at 17:21
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Hint

$$\lim \frac{\frac{\sin(1/n)}{n}}{\frac{1}{n^2}}=1$$

Use the Limit Comparison Test with $\sum_n \frac{1}{n^2}$.

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More generally, if $f$ is any differentiable function such that $f(0) =0$ and $f'(0) \ne 0$, then, for any $c > 0$, $\sum_{k=1}^{\infty} \dfrac{f(1/k)}{k^c} $ converges because $f(1/k) \approx \dfrac{f'(0)}{k}$, so the sum acts like $f'(0)\sum_{k=1}^{\infty} \dfrac1{k^{1+c}}$ which converges.

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    $\begingroup$ I think this result can be generalized with the addition $\hat f'(0)\neq 0$ exists, where $\hat f'(0)=\lim_{x\to 0^+}f'(x)$. $\endgroup$ – Masacroso Feb 25 '17 at 18:14
  • $\begingroup$ I thought of that also, and a further generalization. $\endgroup$ – marty cohen Feb 25 '17 at 20:26
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Following the hint of Simply Beautiful Art

We can say that $\sum_{k=1}^{\infty}\frac{\sin(1/k)}{k} \leq \sum_{k=1}^{\infty}\frac{(1/k)}{k} = \sum_{k=1}^{\infty}\frac{1}{k^2}$

and we know that $\sum_{k=1}^{\infty}\frac{1}{k^2}$ converges thus $\sum_{k=1}^{\infty}\frac{\sin(1/k)}{k}$ converges.

Hope this helps!

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Think it easier: for any $x\in(0,\pi/2)$ we have $\frac{2}{\pi}x\leq \sin(x)\leq x$, hence the series is trivially convergent to something in the interval $\left(\frac{2}{\pi}\zeta(2),\zeta(2)\right)$. The exact value is given by a non-trivial integral: $$ \sum_{k\geq 1}\frac{\sin(1/k)}{k}=\sum_{n\geq 1}\frac{(-1)^{n+1}\zeta(2n)}{(2n-1)!} = \int_{0}^{+\infty}\frac{1}{e^x-1}\sum_{n\geq 1}\frac{(-1)^{n+1}x^{2n-1}}{(2n-1)!^2}\,dx$$ depending on Kelvin's $\text{Bei}_0$ function.

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