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!
Best Answer
Hint
$$\lim \frac{\frac{\sin(1/n)}{n}}{\frac{1}{n^2}}=1$$
Use the Limit Comparison Test with $\sum_n \frac{1}{n^2}$.