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So I want to prove the series given by the seqeunce: $a_n = (\frac{1}{2})^\frac{n+1}{2}$ if $n$ is odd, $a_n = (\frac{1}{3})^{\frac{n}{2}}$ if $n$ is even

converges. I was thinking of comparing it with some series given by the sequence $b_n = 1/2^n + 1/3^n$, but I am not sure if that is legit.

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    Note that $$\left\{\frac{n + 1}{2} : n \in 2\mathbb{N} + 1\right\} = \left\{\frac{n}{2} : n \in 2\mathbb{N}\right\} = \mathbb{N},$$ so you can actually compute your series exactly. – 2017-02-27

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Your series is simply $$ \sum_{n=0}^\infty (1/2)^{\frac{(2n+1)+1}{2}}+\sum_{n=0}^\infty (1/3)^{\frac{2n}{2}}$$ Which we can simplify in the form $$ \frac{1}{2}\sum_{n=0}^\infty (1/2)^{n}+\sum_{n=0}^\infty (1/3)^{n}$$ Both of these are convergent, geometric series.

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Separate the series into two series : $a_n=b_n+c_n$, where $$b_{2p}=0\text{ and }b_{2p+1}=(\frac12)^{p+1},\ c_{2p}=(\frac13)^p\text{ and } c_{2p+1}=0$$ Both series $\sum b_n$ and $\sum c_n$ converge (they are geometric series filled with useless zeroes), and by using the fact that the set of convergent series is a vectorial space, you can conclude $\sum a_n$ converges.