Examining the convergence of $\sum\limits_{n=1}^{+\infty}(-1)^{n+1}\Bigl(1-2\exp\Bigl(\sum\limits_{k=1}^{n}\frac{(-1)^k}k\Bigr)\Bigr)$ Problem: Examine the convergence of the series$$\sum_{n=1}^{+\infty}(-1)^{n+1}\left(1-2\exp\left(\sum_{k=1}^n\frac{(-1)^k}k\right)\right).$$
Solution (part): For the sequence $\alpha_n=(-1)^{n+1}\Big(1-2\exp\Big({\textstyle\sum_{k=1}^{n}\frac{(-1)^k}{k}}\Big)\Big)\,,\; n\in\mathbb{N}\,,$ we have that


*

*$\alpha_{n}>0$ for all $n\in\mathbb{N}$. Indeed
\begin{alignat*}{2}
\log2+\sum_{k=1}^n\frac{(-1)^k}{k}&=\begin{cases}
>0\,, &n\;{\text{even}}\\
<0\,,&n\;{\text{odd}}
\end{cases}\quad&&\Rightarrow\\
\exp\big(\log2+\textstyle{\sum_{k=1}^n\frac{(-1)^k}{k}}\big)&=\begin{cases}
>1\,, & n\;{\text{even}}\\
<1\,,&n\;{\text{odd}}
\end{cases}\quad&&\Rightarrow\\
2\,\exp\big(\textstyle{\sum_{k=1}^n\frac{(-1)^k}{k}}\big)&=\begin{cases}
>1\,, & n\;{\text{even}}\\
<1\,,&n\;{\text{odd}}
\end{cases}\quad&&\Rightarrow\\ 
1-2\,\exp\big(\textstyle{\sum_{k=1}^n\frac{(-1)^k}{k}}\big)&=\begin{cases}
<0\,, & n\;{\text{even}}\\
>0\,,&n\;{\text{odd}}
\end{cases}\quad&&\Rightarrow\\ 
\alpha_n=(-1)^{n+1}\Big(1-2\exp\Big({\textstyle\sum_{k=1}^{n}\frac{(-1)^k}{k}}\Big)\Big)&>0\,.
\end{alignat*}

*$(\alpha_{n})_{n\in\mathbb{N}}$ is decreasing. ( This I wasn't able to prove fully.) 

*$\lim(n\,\alpha_n)\neq0$. Indeed, for all $n\in\mathbb{N}$ we have
\begin{alignat*}{2}
\log2+\sum_{k=1}^{2n}\frac{(-1)^k}{k}&\stackrel{(*)}{>}\log\big(1+\tfrac{1}{20n}\big)>0\quad&&\Rightarrow\\
\exp\big(\log2+\textstyle{\sum_{k=1}^{2n}\frac{(-1)^k}{k}}\big)&>\exp\big(\log\big(1+\tfrac{1}{20n}\big)\big)=1+\frac{1}{20n}\quad&&\Rightarrow\\
-1+2\exp\big(\textstyle{\sum_{k=1}^{2n}\frac{(-1)^k}{k}}\big)&>-1+1+\frac{1}{20n}=\frac{1}{20n}\quad&&\Rightarrow\\
\alpha_{2n}=(-1)^{2n+1}\Big(1-2\exp\big(\textstyle{\sum_{k=1}^{2n}\frac{(-1)^k}{k}}\big)\Big)&>\frac{1}{20n}\quad&&\Rightarrow\\
\lim_{n\to+\infty}(2n\,\alpha_{2n})&\geqslant\lim_{n+\infty}2n\,\frac{1}{20n}=\frac{1}{10}\,.
\end{alignat*}
Because $\alpha_{n}>0$ for all $n\in\mathbb{N}$, we conclude that $\lim(n\,\alpha_n)\neq0$.


By Abel's-Pringsheim's theorem the series diverges.
(*) Proof?
Ideas about the monotonicity of the sequence?
 A: By integrating the geometric series with remainder, we obtain
$$
 - \log 2 = \sum\limits_{k = 1}^n {\frac{{( - 1)^k }}{k}}  + ( - 1)^{n + 1} \int_0^1 {\frac{{x^n }}{{1 + x}}dx} .
$$
Now
\begin{align*}
\int_0^1 {\frac{{x^n }}{{1 + x}}dx} & = \int_0^{ + \infty } {\frac{{e^{ - nt} }}{{e^t  + 1}}dt}  = \frac{1}{{2n}} + \int_0^{ + \infty } {e^{ - nt} \left( {\frac{1}{{e^t  + 1}} - \frac{1}{2}} \right)dt} \\ & = \frac{1}{{2n}} + \int_0^{ + \infty } {e^{ - nt} \mathcal{O}(t)dt}  = \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
\end{align*}
Thus
$$
\sum\limits_{k = 1}^n {\frac{{( - 1)^k }}{k}}  =  - \log 2 + \frac{{( - 1)^n }}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
Consequently,
$$
2\exp \left( {\sum\limits_{k = 1}^n {\frac{{( - 1)^k }}{k}} } \right) = 2\exp \left( { - \log 2} \right)\exp \left( {\frac{{( - 1)^n }}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = 1 + \frac{{( - 1)^n }}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)
$$
and therefore
$$
( - 1)^{n + 1} \left( {1 - 2\exp \left( {\sum\limits_{k = 1}^n {\frac{{( - 1)^k }}{k}} } \right)} \right) = \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
Since the harmonic series diverges, your series is divergent.
