Question

Use root test to determine the series is convergent/divergent or inconclusive

Answer 1

You can first condense the series to make it simpler to study. If n is odd, then n = 2k - 1, and if n is even, then n = 2k for some k ≥ 1. So for each k, you can pair up the k-th terms of the odd- and even-indexed series.

odd:

$\dfrac1{3^{\frac{n+3}2}}=\dfrac1{3^{k+1}}$

even:

$\dfrac1{3^{\frac n2}}=\dfrac1{3^k}$

So the series can be re-indexed as

$\displaystyle\sum_{n=1}^\infty a_n=\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty \left(\frac1{3^{k+1}}+\frac1{3^k}\right)=\sum_{k=1}^\infty\frac4{3^{k+1}}$

By the root test, the series converges, since

$\displaystyle\lim_{k\to\infty}\sqrt[k]{\left|\frac4{3^{k+1}}\right|}=\frac13\lim_{k\to\infty}\left(\frac43\right)^{\frac1k}=\frac13<1$