Discuss if the Series is absolutely convergent or conditional convergent Discuss if the Series $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p+(-1)^{n-1}},p\gt\frac{1}{2}$$
is absolutely convergent or conditional convergent.
It's easy to see $$|\frac{(-1)^{n-1}}{n^p+(-1)^{n-1}}|=\frac{1}{n^p+(-1)^{n-1}}\ge \frac{1}{n^p-1} $$
Hence $$\sum_{n=1}^{\infty} |\frac{(-1)^{n-1}}{n^p+(-1)^{n-1}}|$$
is divergent. And how consider the convergence and divergence $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^p+(-1)^{n-1}},p\gt\frac{1}{2}$$
 A: The series is conditionally convergent if $1 \geqslant p > 1/2$.  Absolute convergence for $p > 1$ follows from the limit comparison test with the convergent series $\sum n^{-p}\,(p > 1).$
Note that
$$\sum_{n=2}^m\frac{(-1)^{n-1}}{n^p+(-1)^{n-1}} = \sum_{n=2}^m\frac{(-1)^{n-1}}{n^p+(-1)^{n-1}}\frac{n^p-(-1)^{n-1}}{n^p-(-1)^{n-1}} \\ = \sum_{n=2}^m\frac{(-1)^{n-1}n^p}{n^{2p}-1} - \underbrace{\sum_{n=2}^m\frac{1}{n^{2p}-1}}_{\text{convergent } \iff p > 1/2} $$
Now you can show that $n^p/(n^{2p} - 1)$ is decreasing for $p > 0$ and the first sum on the RHS converges by the AST.  However, the second sum diverges for $0 < p \leqslant 1/2$.
A: Basic idea.  Combine terms in pairs an see if that is absolutely. convergent.  To simplify I'll ignore $(-1)^{n-1}$ in the denominator.  Consecutive pairs add up (in magnitude to) $|\frac{1}{n^p}-\frac{1}{(n+1)^p}|<\frac{(n+1)^p-n^p}{n^{2p}}=q(n)$  The numerator $=n^p((1+\frac{1}{n})^p-1)=n^p(\frac{p}{n}+O(\frac{1}{n^2}))\approx pn^{p-1}$.  Therefore $q(n)\approx \frac{p}{n^{p+1}}$.  For $p\gt 0$, the series of pairs is absolutely convergent, so the original series is conditionally convergent.   
