Determine the value of $p\ge0$ for which the series $$\sum_{n=1}^\infty\frac{n(-1)^n}{(1+n^2)^p}$$ is divergent, conditionally convergent or absolutely convergent.
I have done the absolute convergent part. $$\left|\frac{n(-1)^n}{(1+n^2)^p}\right|\le \left|\frac{n}{(1+n^2)^p} \right|\le\left| \frac{1}{(1+n^2)^p}\right|\le\left|\frac{1}{n^{2p}} \right|$$ Hence, by comparison test, the series converges absolutely when $p>\frac{1}{2}$.
Is this series diverges when $p\le\frac{1}{2}$? I have no idea how to show that for what $p$ it will converges conditionally. Thanks for any comments.
Best Answer
Absolute convergence comes from $$\left|\frac{n(-1)^n}{(1+n^2)^p}\right| \sim \frac{1}{n^{2p-1}}$$ so you need $2p-1>1$.
Conditional convergence we will have when $2p-1>0$
Addition.
For conditional convergence we need $\frac{n}{(1+n^2)^p} \to 0$ monotonously. Possibly most easy way to prove is consider $f(x)=\frac{x}{(1+x^2)^p}$ and calculate $f'(x)=\frac{1+x^2(1-2p)}{(1+x^2)^{p+1}}$, from where, for $2p-1>0$ we obtain desired.
On another hand if there is some reason to want solve inequality $\frac{n+1}{(1+(n+1)^2)^p} \leqslant \frac{n}{(1+n^2)^p}$ straight ahead, then we can write $(n+1)(1+n^2)^p \leqslant n(1+(n+1)^2)^p$ and notice, that while maximal power of $n$ on both sides is $n^{2p+1}$ with same coefficient $1$, then member $n^{2p}$ on left side have coefficient $1$ and on right side coefficient $2p$. So, again, we obtain decreasing when $2p-1>0$.