How to judge the positive series$\frac{\left[2+(-1)^n\right]^n}{n}\left(\frac{1}{3}\right)^n$ when comparision/root test don't work?

Question

Comparision/root test don't work for the serise$\frac{\left[2+(-1)^n\right]^n}{n}\left(\frac{1}{3}\right)^n$

Answer 1

Let $a_n= \frac{(2+(-1)^n)^n}{n} \left( \frac{1}{3} \right)^n$. You want to see whether we can apply comparison test can be applied to the series $\sum_{n=1}^{\infty} a_n$.
Define $b_n = a_n$ whenever $n$ is even and $b_n = 0$ when $n$ is odd. Clearly $0 \le b_n \le a_n$ for all $n \in \mathbb{N}$. Observe that when $n$ is even, $b_n = a_n = \frac{1}{n}$.
We know that the series $\sum_{n=1}^{\infty} \frac{1}{2n}$ diverges. Thus, $\sum_{n=1}^{\infty} b_n$ diverges.
Since $0 \le b_n \le a_n$, we apply comparison test to conclude that $\sum_{n=1}^{\infty} a_n$ diverges.