I'm aware that the alternating p-series $\sum_{n=1}^{\infty}(-1)^n1/n^p$ converges absolutely for $p>1$ and conditionally for $0<p\leq1$, and these are fairly straightforward to prove. But what about $p\leq0$, where the alternating series test fails?
I know that the series cannot be absolutely convergent, since $\sum_{n=1}^{\infty}|(-1)^n1/n^p|=\sum_{n=1}^{\infty}1/n^p$ and this is just a p-series with $p\leq0$ which diverges.
Metamorphy already answered your question. But adding more details: the answers is no because the terms of your series are $$a_n=(-1)^nn^p$$ with $p\ge0$ and $a_n$ does not converges to $0$. This implies that your series diverges.