Show that the series $\sum_{n=0}^\infty (-1)^n \frac{1}{(n+1)^p}$ converges and converges absolutely

Question

Show that the series

$\sum_{n=0}^\infty (-1)^n \frac{1}{(n+1)^p}$

converges for $p > 0$ and converges absolutely for $p > 1$.

I'm a little confused on this. The fact that it's an alternating series is throwing me off.. should I be comparing it to the $p$-series of $\sum_{n=1}^\infty \frac{1}{n^p}$ ? I know that series converges by the comparison test, and if $p \le 1$ it diverges by comparison with the harmonic series since $\frac{1}{n^p} \ge \frac{1}{n}$.

Or should I be using a different method all together?

Answer 1

You will most likely want to use two different methods for each part of the problem; you have been asked to prove two different statements, so it is not unreasonable to expect to use two methods.

For the first statement, that you have convergence when $p>0$, do you recall the Alternating Series Test?

For the second statement, that you have absolute convergence when $p>1$, you should write down the definition of what absolute convergence means, and then attempt to use the comparison test (maybe with a change of variables if you don't like the indexing) in order to get a proof.