Does parity matter for $\lim_{n\to \infty}\left(\ln 2 -\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots -\frac{(-1)^n}{n}\right)\right)^n =\sqrt{e}$?
$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$$ is well-known to tend to $\log 2$, and the expression inside the main parenthesis oscillates around $1$. One can expect an asymptotic behavior like
$$1\pm\frac1{2n}.$$
Then taking the $n^{th}$ power, the value will alternatively rejoin $e^{1/2}$ and $e^{-1/2}$, so the limit of the sequence does not exist.
More precisely, if we group the terms in pairs, we have alternatively
$$S_{2n}=1+\sum_{k=2n+2}^\infty\frac1{2k(2k+1)}\sim 1+\frac1{4n}$$
and
$$S_{2n+1}=1+\sum_{k=2n+2}^\infty\frac1{2k(2k+1)}-\frac1{2n+1}\sim 1-\frac1{4n},$$ approximating the sums by integrals.
Taking the power, we have
$$S_{2n[+1]}^{2n}\sim\left(1\pm\frac1{4n}\right)^{2n}\sim e^{\pm1/2}.$$
Preliminaries
Note that $$ \begin{align} \frac12\left(\frac1{2k}-\frac1{2k+2}\right) \le\frac1{2k}-\frac1{2k+1} \le\frac12\left(\frac1{2k-1}-\frac1{2k+1}\right)\tag1 \end{align} $$ Summing $(1)$ for $k\ge n$ gives $$ \frac1{4n}\le\sum_{k=n}^\infty\left(\frac1{2k}-\frac1{2k+1}\right)\le\frac1{4n-2}\tag2 $$ Furthermore, $$ \begin{align} \frac12\left(\frac1{2k+1}-\frac1{2k+3}\right) \le\frac1{2k+1}-\frac1{2k+2} \le\frac12\left(\frac1{2k}-\frac1{2k+2}\right)\tag3 \end{align} $$ Summing $(3)$ for $k\ge n$ gives $$ \frac1{4n+2}\le\sum_{k=n}^\infty\left(\frac1{2k+1}-\frac1{2k+2}\right)\le\frac1{4n}\tag4 $$
Two Limits
Inequality $(2)$ gives $$ \begin{align} \log(2)+\sum_{k=2}^{2n-1}\frac{(-1)^k}k &=1-\sum_{k=2n}^\infty\frac{(-1)^k}k\tag5\\ &=1-\sum_{k=n}^\infty\left(\frac1{2k}-\frac1{2k+1}\right)\tag6\\ &=1-\left[\frac1{4n},\frac1{4n-2}\right]\tag7 \end{align} $$ where $[a,b]$ is a number between $a$ and $b$.
Likewise, inequality $(4)$ gives $$ \begin{align} \log(2)+\sum_{k=2}^{2n}\frac{(-1)^k}k &=1-\sum_{k=2n+1}^\infty\frac{(-1)^k}k\tag8\\ &=1+\sum_{k=n}^\infty\left(\frac1{2k+1}-\frac1{2k+2}\right)\tag9\\ &=1+\left[\frac1{4n+2},\frac1{4n}\right]\tag{10} \end{align} $$ Therefore, $(7)$ says that for an even number of terms in the sum $$ \begin{align} \lim_{n\to\infty}\left(\log(2)+\sum_{k=2}^{2n-1}\frac{(-1)^k}k\right)^{2n-1} &=\lim_{n\to\infty}\left(1-\left[\frac1{4n},\frac1{4n-2}\right]\right)^{2n-1}\tag{11}\\[6pt] &=e^{-1/2}\tag{12} \end{align} $$ and $(10)$ says that for an odd number of terms in the sum $$ \begin{align} \lim_{n\to\infty}\left(\log(2)+\sum_{k=2}^{2n}\frac{(-1)^k}k\right)^{2n} &=\lim_{n\to\infty}\left(1+\left[\frac1{4n+2},\frac1{4n}\right]\right)^{2n}\tag{13}\\[6pt] &=e^{1/2}\tag{14} \end{align} $$
Conclusion
Using $(11)$, $(13)$, and the inequality $$ e^{\frac x{1+x}}\le1+x\le e^x\tag{15} $$ we get $$ \begin{align} \left(\log(2)+\sum_{k=2}^n\frac{(-1)^k}k\right)^n &=\left(1+(-1)^n\left[\frac1{2n+2},\frac1{2n}\right]\right)^n\tag{16}\\ &=\left\{\begin{array}{} e^{\frac12-\left[0,\frac3{4n+6}\right]}&\text{if $n$ is even}\\ e^{-\frac12+\left[-\frac1{4n-2},\frac1{2n+2}\right]}&\text{if $n$ is odd} \end{array}\right.\tag{17} \end{align} $$ Therefore, the limit does not exist, but if we restrict $n$ to be even or $n$ to be odd, then each of those limits do exist.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\, } \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[15px,#ffd]{\lim_{n \to \infty}\braces{\ln\pars{2} - \bracks{-\,{1 \over 2} + {1 \over 3} - {1 \over 4} + \cdots -{\pars{-1}^{n} \over n}}}^{n} = \root{\expo{}}}:\ {\Large ?}}$
\begin{align} &\bbox[15px,#ffd]{\lim_{n \to \infty}\braces{\ln\pars{2} - \bracks{-\,{1 \over 2} + {1 \over 3} - {1 \over 4} + \cdots -{\pars{-1}^{n} \over n}}}^{n}} \\[5mm] = &\ \lim_{n \to \infty}\bracks{\ln\pars{2} + \sum_{k = 2}^{n}{\pars{-1}^{k} \over k}}^{n} \\[5mm] = &\ \lim_{n \to \infty}\bracks{\ln\pars{2} + 1 + \sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k} - \sum_{k = n + 1}^{\infty}{\pars{-1}^{k} \over k}}^{n} \\[5mm] = &\ \lim_{n \to \infty}\bracks{1 - \sum_{k = n + 1}^{\infty}{\pars{-1}^{k} \over k}}^{n} = \lim_{n \to \infty}\bracks{1 - \pars{-1}^{n + 1}\sum_{k = 0}^{\infty} {\pars{-1}^{k} \over k + n + 1}}^{n} \\[5mm] = &\ \lim_{n \to \infty}\bracks{1 + \pars{-1}^{n}\sum_{k = 0}^{\infty}\pars{% {1 \over 2k + n + 1} - {1 \over 2k + n + 2}}}^{n} \\[5mm] = &\ \lim_{n \to \infty}\bracks{1 + {1 \over 4}\pars{-1}^{n}\sum_{k = 0}^{\infty} {1 \over \pars{k + n/2 + 1/2}\pars{k + n/2 + 1}}}^{n} \\[5mm] = &\ \lim_{n \to \infty}\braces{1 + {1 \over 2}\pars{-1}^{n} \bracks{\Psi\pars{{n \over 2} + 1} - \Psi\pars{{n \over 2} + {1 \over 2}}}}^{n} \end{align} Note that \begin{align} &\Psi\pars{{n \over 2} + 1} - \Psi\pars{{n \over 2} + {1 \over 2}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {1 \over n} - {1 \over 2n^{2}} \end{align}
Then, \begin{align} &\mbox{} \\ &\bbx{\bracks{\ln\pars{2} + \sum_{k = 2}^{n}{\pars{-1}^{k} \over k}}^{n} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, \bracks{1 + \pars{-1}^{n}{\color{red}{1/2} \over n}}^{n}} \\ & \end{align}