The convergence of $\sum\frac{(-1)^n}{n^a+(-1)^n}$ 
Test the following series for convergence:
  $$\sum\frac{(-1)^n}{n^a+(-1)^n} , \quad a> 0.$$ 

I tried to use $$\frac{U_{n+1}}{Un}.$$
I get $$\frac{-n^a-(-1)^n}{(n+1)^a+(-1)^{n+1}},$$ 
but the limit is undefined.  
$\dfrac{1}{(n+1)^a+(-1)^n}$
 is not decreasing, so I can't use Leibniz test.
 A: Hint: Think of the sum as 
$$\sum_{k=1}^\infty\left (\frac{1}{(2k)^a + 1} - \frac{1}{(2k+1)^a - 1}\right).$$
A: Observe that
$$\sum_{n=2}^m \frac{(-1)^n}{n^a + (-1)^n} = \sum_{n=2}^m \frac{(-1)^n\,[n^a - (-1)^n]}{[n^a + (-1)^n]\,[n^a - (-1)^n]} \\ =  \underbrace{\sum_{n=2}^m \frac{(-1)^n\,n^a}{n^{2a} - 1}}_{\text{converges for }\, a \,> \, 0} -  \underbrace{\sum_{n=2}^m \frac{1}{n^{2a} - 1}}_{\text{converges for }\, a \,> \, 1/2} $$
Convergence of the first sum on the RHS can be established by the alternating series test.  The second sum can be compared with a p-series that converges if $a > 1/2$ and diverges if $a \leqslant 1/2$. Hence, the original series with $a > 0$ diverges for $0 < a \leqslant 1/2$ and converges for $a > 1/2$.
A: The infinite sum can be expressed as the sum of two infinite sums, one for odds and one for evens. Further I will assume that all $n$ are positive. I will explain how there is a similar problem with an equivalent solution at the end.
NOTE by original author: This solution is incorrect, I'll work on it. To separate into two series, the original must be convergent.
\begin{equation}
\sum\frac{(-1)^n}{n^a+(-1)^n}=\sum\frac{1}{(2n)^a+1} - \sum\frac{1}{(2n+1)^a-1}~~~~~~~~~~~~~~~~~(1)
% if someone has a better way to space that eqn number, please do
\end{equation}
These series can show convergence if a>1 when compared to similar series not including the constant $\pm1$ using the limit comparison test. The limit comparison test states that if the limit of the ratio of the two series is non-zero and finite, then the two series must both diverge or both converge. The limits are evaluated using L'Hopital's rule.
$$
\lim_{n\to\infty}\frac{(2n)^a}{(2n)^a+1} = \lim_{n\to\infty}\frac{n^{a-1}}{n^{a-1}}=1
$$
similarly for the other series.
$$
\lim_{n\to\infty}\frac{(2n+1)^a}{(2n+1)^a-1} = \lim_{n\to\infty}\frac{(2n+1)^{a-1}}{(2n+1)^{a-1}}=1
$$
Both series $\sum(2n)^{-a}$ and $\sum(2n+1)^{-a}$ can be shown to converge for $a>1$ by integral test. So the both RHS series in Eq $1$ are convergent and is equal to the difference of two finite numbers, which means your series is convergent for $a>1$.
As far as using $n<0$ any $n^a$ can be expressed as $(-k)^a$ for $k>0$. Then using Euler's identity for $-1$, 
$$
(-k)^a = (e^{i~\pi/2}k)^a = e^{ia~\pi/2}k^a
$$
and the original sum can be expressed as follows for some complex constant, $C$ and it would cancel in the limit calculation
$$
\sum_{k>0}\frac{(-1)^k}{Ck^a+(-1)^k}
$$
Just realized I was too slow.
