<s>[INST] Solution Verification: $\sum_{n=1}^{\infty}2^{-n}\left(1+\frac{(-1)^n}{n}\right)^{n^2}$ I don't have the 'official' solution to this series so I wanted to verify mine.
\begin{equation*}
(s_n)=\sum_{n=1}^{\infty}2^{-n}\left(1+\frac{(-1)^n}{n}\right)^{n^2}
\end{equation*}
$\underline{\text{Claim:}}$ $(s_n)$ diverges.
$\underline{\text{Proof:}}$ Using root test:
\begin{equation*}
\sqrt[n]{\left|a_n\right|}=\sqrt[n]{\left|2^{-n}\cdot\left(1+\frac{(-1)^n}{n}\right)^{n^2}\right|}=\sqrt[n]{2^{-n}\cdot\left(1+\frac{(-1)^n}{n}\right)^{n^2}}.
\end{equation*}
Further:
\begin{equation*}
\sqrt[n]{2^{-n}\cdot\left(1+\frac{(-1)^n}{n}\right)^{n^2}}=\sqrt[n]{2^{-n}}\cdot\sqrt[n]{\left(1+\frac{(-1)^n}{n}\right)^{n^2}}=\frac{1}{2}\cdot\left(1+\frac{(-1)^n}{n}\right)^n.
\end{equation*}
\begin{equation*}
\lim_{n\rightarrow\infty}\frac{1}{2}\cdot\left(1+\frac{(-1)^n}{n}\right)^n=\begin{cases}\frac{e}{2} & \text{n is even,} \\ \frac{1}{2e} & \text{n is odd.}
\end{cases}
\end{equation*}
Therefore:
\begin{equation*}
C=\limsup\sqrt[n]{\left|a_n\right|}=\frac{e}{2}.
\end{equation*}
Since $C=\frac{e}{2}>1$: $~~~(s_n)$ diverges.
 [/INST] Yes. This works.
More directly, we note that, for $n\geq 1,$ $$\left(1+\frac 1 n\right)^n\geq 2$$
This follows by noting the first two terms of the binomial expansion of the left side are $1+n\cdot \frac1n=2.$
Then for even $n$:
$$2^{-n}\left(1+\frac{(-1)^n}n\right)^{n^2}\geq 2^{-n}\cdot 2^n=1.$$
So the terms do not converge to $0.$
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