For which parameter values does the series $\sum 1/(n^\beta + (-1)^n n^\alpha)$ converge? Given that $0 < \beta < \alpha \le 1$, for which $\alpha,\beta$ does the series
$\sum^\infty_{n=1}\frac {1}{n^\beta + (-1)^n n^\alpha}$ 
converge ?

Write $a_n$ for the general term of the series. Also, let  $\delta = a-b$.
Set 
$$b_n = a_n - \frac{(-1)^n}{ n^{a}} = \frac{(-1)^n}{n^a (1+ (-1)^n n^{-\delta})}-\frac{(-1)^n}{n^a}=\frac{(-1)^n}{n^a} \times \left (1- \frac{1}{1+ (-1)^n n^{-\delta}}\right)=(*).$$
Now $$(*)= \frac{(-1)^n}{n^a} \times \frac{(-1)^n n ^{-\delta}}{1+(-1)^n n^{-\delta}}=\frac{1}{n^{a+\delta}}\times \frac{1}{1+(-1)^n n^{-\delta}}.$$
Therefore $\sum b_n$ is a positive series, and by comparison converges if and and only if $$a+\delta =2a-b >1.$$
Since  $\sum (-1)^n /n^a$ converges for all $a>0$, and $a_n = b_n + (-1)^n/n^a$, it follows that  $\sum a_n $ converges if and only if $\sum b_n$ converges, that is if and only if $2a-b>1$. 
