Does parity matter for $\lim_{n\to \infty}\left(\ln 2 -\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots -\frac{(-1)^n}{n}\right)\right)^n =\sqrt{e}$? 
Prove that $$\lim_{n\to \infty}\left(\ln 2 -\left(-\frac12+\frac13-\frac14+\cdots  -\frac{(-1)^n}n\right)\right)^n =\sqrt{e}$$

I happened to encounter this problem proposed  by Mohammed Bouras,Morocco in the facebook group of Romanian mathematical Magazine
As per the  title,  I think the limit of the problem depends upon the parity of $n$. That is,if $n$ is even, the limit is $\frac1{\sqrt e}$ otherwise as stated.
My query is, Does the parity indeed matters for this problem? And if it matters what should be  the conclusion for limit of the problem?
Here is my try
we will show that the there exist two different limits  for above problem.
For  $0< x\leq 1$, we define the functions $$f(x)=\ln(1+x),\; \displaystyle g(x)=\sum_{k=1}^n \frac{(-x)^k}{k+1}$$ and we note that $$\begin{aligned}f(x)-g(x) &= x-\sum_{k=2}^\infty(-1)^{k+n} \frac{x^{k+n}}{k+n}\\&=x+\sum_{k=2}^{\infty} (-1)^{k+n} \int_0^x t^{k+n-1}dt\\&=x+(-1)^n\int_0^x t^n\left(\sum_{k=1 }^\infty(-1)^k t^{k-1} \right)dx\\&=x-(-1)^n\int_0^x\frac{t^n}{1+t} dt\end{aligned}$$ hence for $x=1$   we have then $$f(1)-g(1)=\ln(2)-\sum_{k=1}^\infty\frac{(-1)^k}{k+1}=1-(-1)^n\int_0^1\frac{t^n}{1+t}dt$$ Note that latter integral is know result however,here we shall derive it and  we shall show that

$$\displaystyle\lim_{n\to\infty}(f(1)-g(1))^n =\begin{cases}\sqrt{e}\; \text{if }  \, n\in 2n-1 \\  \frac1{\sqrt{e}} \; \text{otherwise}\end{cases}$$

We solve the following integral for any $n>0$. By  polynomial long division it  is trivial to note that  $$\int_0^1\frac{t^n}{t+1}dt=(-1)^n\int_0^1\left(\frac{1}{t+1}-\sum_{0\leq j\leq n}(-1)^j t^{j-1}\right)dt$$  and hence on integrating $\displaystyle \int_0^1\frac{t^n}{1+t}dt$ $$\begin{aligned}&=(-1)^n\left(\log(2) -\sum_{1\leq j\leq n} \frac{(-1)^{j+1}}{j}\right)\\&=2^{-1}\left(-\psi\left(\frac{n+1}2\right)+\psi\left(\frac{2n+1}2\right)\right)\\&=\frac12\left(H_{\frac{n}2}-H_{\frac{n-1}2}\right)\end{aligned}$$ Further we note that $H_n\approx \gamma +\ln n +\frac1{2n}-O(n^{-2})$  with which we deduce that $$H_{\frac{n}2} -H_{\frac{n-1}2} \approx  \frac1n-\ln\left(\frac{n-1}n\right)+\frac1{n-1}$$ for all $n>1$ and hence $H_{\frac{n}{2}} -H_{\frac{n-1}2} \to \frac1n$as $n$ gets larger. Thus we have for $$\lim_{n\to\infty}(f(1)-g(1))^n= \lim_{n\to\infty} \left(1-\frac{(-1)^n}{2n}\right)^n=e^{-\frac{(-1)^n}2} =\sqrt{e^{-(-1)^n}}$$
therefore  if $n$ is even we have limit as $\displaystyle \frac1{\sqrt{e}}$ and if $n$ is odd we  have limit $ \displaystyle \sqrt{e}$.
Since we have two different limits. Does it have limit ?
Thank you
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\, }
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[15px,#ffd]{\lim_{n \to \infty}\braces{\ln\pars{2} -
\bracks{-\,{1 \over 2} + {1 \over 3} - {1 \over 4} + \cdots -{\pars{-1}^{n} \over n}}}^{n} = \root{\expo{}}}:\ {\Large ?}}$

\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}
A: 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
$$S_{2n[+1]}^{2n}\sim\left(1\pm\frac1{4n}\right)^{2n}\sim e^{\pm1/2}.$$
