How can I show that this series diverges? $$ \sum_{n=1}^{\infty} \frac{\bigl(2 + (-1)^n\bigr)^n}{\sqrt{n}\,3^n} $$
The standard tests kinda fail in this case. I am not getting any clue on how to proceed. How can I use the partial sum-boundedness to prove it or the $n$th partial sum?
For every $n \geq 1$, let $u_n = \dfrac{\bigl(2 + (-1)^n\bigr)^n}{\sqrt{n}\,3^n}$.
Then one has $$u_{2n} = \dfrac{3^{2n}}{\sqrt{2n}3^{2n} }= \dfrac{1}{\sqrt{2n}}$$
and $$u_{2n+1} = \dfrac{1}{(\sqrt{2n+1})3^{2n+1}}$$
Now it is easy to see that the series $\displaystyle \sum u_{2n}$ diverges whereas the series $\displaystyle \sum u_{2n+1}$ converges, and hence, by sum, the series $\displaystyle \sum u_{n}$ diverges.