Does parity matters ?

I wish to share my one of those two proposed solutions. If any errors or typo are spotted in the solution I would be glad to be corrected.

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 $f(x)-g(x) = x-\sum_{k=2}^{\infty}(-1)^{k+n} \frac{x^{k+n}}{k+n}=x+\sum_{k=2}^{\infty} (-1)^{k+n} \int_0^x t^{k+n-1}dt$ $=x+(-1)^n\int_0^x t^n\left(\sum_{k=1 }^{\infty}(-1)^k t^{k-1} \right)dx=x-(-1)^n\int_0^x\frac{t^n}{1+t} dt$ hence for $x=1$ we have then $f(1)-g(1)=\ln(2)-\sum_{k=1}^{\infty}\frac{(-1)^k}{k+1}=1-(-1)^n\int_0^1\frac{t^n}{1+t}dt$ Now we shall show that $\displaystyle \lim_{n\to\infty}(f(1)-g(1))^n =\begin{cases} \sqrt{e}\; \text{if } \, n\in 2n-1 \\ \frac{1}{\sqrt{e}} \; \text{otherwise}\end{cases}$

We solve the following integral for any $0\leq m\leq n$ . By polynomial long division it is trivial to note that $\int_{0}^1\frac{t^{m}}{t+1}dt=(-1)^m\int_0^1\left(\frac{1}{t+1}-\sum_{0\leq j\leq m}(-1)^j t^{j-1}\right)dt$ and hence on integrating $\displaystyle \int_0^1\frac{t^{m}}{1+t}dt$ $=(-1)^m\left(\log(2) -\sum_{1\leq j\leq k} \frac{(-1)^{j+1}}{j}\right)=2^{-1}\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{2n+1}{2}\right)\right)=\frac{1}{2}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right)$ Further we note that $H_n\approx \gamma +\ln n +\frac{1}{2n}-O(n^2)$ with which we deduce that $H_{\frac{n}{2}} -H_{\frac{n-1}{2}} \approx \frac{1}{n}-\ln\left(\frac{n-1}{n}\right)+\frac{1}{n-1}$ for all $n>1$ and hence $H_{\frac{n}{2}} -H_{\frac{n-1}{2}} \to \frac{1}{n}$ 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 $\frac{1}{\sqrt{e}}$ and if $n$ is odd we have limit $\sqrt{e}$ .

As $\alpha =2$ and $l_1=\frac{1}{e}$ and $l_2=\sqrt{e}$ . Finally $\sum_{1\leq j\leq \alpha} \left(\sum_{1\leq i\leq k} l_i\right)^{j} =\sqrt e +\frac{1}{\sqrt{e}}+\left(\sqrt{e}+\frac{1}{\sqrt{e}}\right)^2\approx 7.3414.$