It is supposed to converge by the comparison test... Just anyone is able to explain me how? I'm getting crazy with this particular one...
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to domy7997, you have very poor penmanship. Those are clearly m's in the image you posted.
It may serve better if you used some devise other than a mobile phone.
Also be reminded that this is an English language website.
Now just to be sure that the series is: $\displaystyle \sum\limits_{n = 1}^\infty {n!{{\left[ {\log \left( {1 + \frac{2}{n}} \right)} \right]}^n}} $
If that is correct then the ratio test is a better way to prove convergence.
we could use
$\displaystyle \log \left(1+\frac{2}{n}\right)<\frac{2}{n}$
$\displaystyle n!\left(\log \left(1+\frac{2}{n}\right)\right)^n<\frac{2^nn!}{n ^n}$
and compare with the series
$\displaystyle \sum _{n=1}^{\infty } \frac{2^nn!}{n^n}$
which converges by the $n^{th}$ root test using
$\displaystyle \lim_{n\to \infty } \left(\frac{\sqrt[n]{n!}}{n}\right)=\frac{1}{e}$
The proof is over a page long when done right. So all I will do is a brief outline.
$\displaystyle \log \left[ {(N - 1)!} \right] \le \sum\limits_{k = 2}^N {\log (k - 1)} \le \int_1^N {\log (x)} = N\log (N) - N + 1$
Raise to power of $e$
$\displaystyle \begin{align*}(N-1)!&\le \dfrac{N!}{e^N}e \le N!\\\dfrac{e^N}{N}&\le\dfrac{N^N}{N!}e\le e^N\\\dfrac{e}{\sqrt[N]{N}}&\le\sqrt[N]{\dfrac{N^N}{N!}}\sqrt[N]e\le e\end{align*}$Now if $N\to\infty$ then if follow that $\sqrt[N]{\dfrac{N^N}{N!}}\to e$.
Just take the reciprocal.