This is a question we asked on a second semester calculus test.
For what values of $p$ does this series converge?
$$\sum_{n=1}^{\infty}\frac{\sin(1/n)}{n^p}$$
I believe that it actually can be shown that $p> 0$ is a valid answer.
However. I am interested in finding a proof that is simple enough that a beginning calculus student could do on their own cognition.
Is there a simple way to give the exact values of $p$ for which the series converges?
Best Answer
Limit comparison test.
$$\lim \frac{\sin(1/n)/n^p}{1/n^{p+1}}=1,$$
$\sum\frac{1}{n^{p+1}}$ converges when $p>0$, diverges when $p\leq 0$.