Two series questions I'm trying to correct an old quiz.  I want to see if I have sufficiently corrected the second problem.  The first I am still a bit unsure what to do.
(1)  Show that the following series is divergent if $\alpha \in \mathbb{R}$ such that $|\alpha|<1$.
$$\sum_{k=1}^{\infty} \frac{1}{1+\alpha^k}$$
The above I wasn't sure what to do so I left it blank.
(2)  Use the root test to decide whether or not the following series converges:
$$\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+...$$
First note: $a_n =\begin{cases} \frac{1}{2^n} \text{ if n is odd}\\ \frac{1}{3^n} \text{ if n is even } \end{cases}$.
So $\liminf a_n = \sqrt[2n]{\frac{1}{3^n}}=\frac{1}{3}$ and $\limsup a_n = \sqrt[2n-1]{\frac{1}{2^n}}=1$  So the sequence diverges.  I just want to make sure I got the values correct on this one.

Note that if $\alpha \in (-1,1)$, then $1 + \alpha^k \leq 2$ so that
$$
\frac{1}{1 + \alpha^k} \geq \frac{1}{2}.
$$
For the second one, since $\frac{1}{3} \leq \frac{1}{2}$, then:
$$
\frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^3} + \dots \leq \sum_{k=1}^{\infty}  \frac{1}{2^k} = 1
$$
If you want to use the root test: Note that 
\begin{align}
\limsup_{n \to \infty} \sqrt[n]{a_n} = \limsup_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = \frac{1}{2} < 1
\end{align}
so the series converges absolutely.
