Convergence of $\sum_{n=0}^\infty\frac{2}{\alpha^n+\alpha^{-n}}$ 
Determine for which values of the parameter $\alpha\in\mathbb{R}$ ($\alpha\neq0$) the following series is convergent:
  $$\sum_{n=0}^\infty\frac{2}{\alpha^n+\alpha^{-n}}$$

I have developed the following cases:
· $\alpha>1,\ \alpha^{-n}<\alpha^n,\ \sum_{n=0}^\infty\frac{2}{\alpha^n+\alpha^{-n}}\leq\sum_{n=0}^\infty\frac{2}{2\alpha^n}=\sum_{n=0}^\infty\frac{1}{\alpha^n}$, convergent as $|1/\alpha|<1$
· $\alpha=1,\ \sum_{n=0}^\infty\frac{2}{1+1}=\sum_{n=0}^\infty1=\infty$, divergent
· $0<\alpha<1,\ \alpha^{-n}>\alpha^n,\ \sum_{n=0}^\infty\frac{2}{\alpha^n+\alpha^{-n}}\leq\sum_{n=0}^\infty\frac{2}{2\alpha^{-n}}=\sum_{n=0}^\infty\alpha^n$, convergent as $|\alpha|<1$
I have certain doubts about what happens when $\alpha<0$. This is what I have done:
· $\alpha<0,\ \sum_{n=0}^\infty\frac{2}{\alpha^n+\alpha^{-n}}=\sum_{n=0}^\infty(-1)^n\frac{2}{|\alpha|^n+|\alpha|^{-n}}$
Now I try to apply alternating series test with $a_n=\frac{2}{|\alpha|^n+|\alpha|^{-n}}$. We have that $\lim a_n=0$, and to show that this sequence is decreasing, $a_n'=\frac{2(|\alpha|^n\ln(|\alpha|)-|\alpha|^{-n}\ln(|\alpha|))}{(positive\ things)}=\frac{2\ln(|\alpha|)(|\alpha|^n-|\alpha|^{-n})}{(positive\ things)}$.
Now we have that, if $-1<\alpha<0$, $\ln(|\alpha|)<0$ and $(|\alpha|^n-|\alpha|^{-n})<0$, so the derivative is positive.
If $\alpha<-1$, $\ln(|\alpha|)>0$ and $(|\alpha|^n-|\alpha|^{-n})>0$, and the derivative is also positive, so alternating series test does not work. How should I test convergence with $\alpha<0$?

As noticed in the comment recall that "absolute convergence implies convergence", then for $\alpha\neq-1$
$$\sum_{n=0}^\infty \left|(-1)^n\frac{2}{|\alpha|^n+|\alpha|^{-n}}\right|=\sum_{n=0}^\infty\frac{2}{|\alpha|^n+|\alpha|^{-n}}\le \infty $$
and for $\alpha=-1$
$$\sum_{n=0}^\infty(-1)^n\frac{2}{|\alpha|^n+|\alpha|^{-n}}=\sum_{n=0}^\infty(-1)^n$$
