Does parity matter for $\lim_{n\to \infty}\left(\ln 2 -\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots -\frac{(-1)^n}{n}\right)\right)^n =\sqrt{e}$?

The alternating series

$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}n$$ 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.