Does the series $\sum_{k=1}^\infty\frac{\sin(1/k)}{k}$ converge? 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!
 A: Hint
$$\lim \frac{\frac{\sin(1/n)}{n}}{\frac{1}{n^2}}=1$$
Use the Limit Comparison Test with $\sum_n \frac{1}{n^2}$.
A: 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.
A: 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!
A: 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.
