Hi
i solve a problem about the convergence/divergence of a sequence ..
and i use the trick which in the attachment ..
is my solution correct ?
Yes that is true.
So for any spectators what is going on is that TWiX is just making the observation that if $\displaystyle \sum_{n\in\mathbb{N}}\sigma_n$ converges it is necessary for $\displaystyle \lim_{n\to\infty}\sigma_n$. And of course there are a lot of ways to prove that $\displaystyle \sum_{n\in\mathbb{N}}\sigma_n$ converges thaty may be easier than actually computing $\displaystyle \lim_{n\to\infty}\sigma_n$.
Example: Compute $\displaystyle \lim_{n\to\infty}\frac{(n!)^2}{(2n)!}$
Solution: Note that $\displaystyle \lim_{n\to\infty}\left|\frac{\left[(n+1)!\right]^2}{(2(n+1))!}\cdot\frac{(2n)!}{(n!)^2}\right|$$\displaystyle =\lim_{n\to\infty}\frac{(n+1)^2(n!)^2}{(2n+2)(2n+1 )(2n)!}\cdot\frac{(2n)!}{(n!)^2}=\lim_{n\to\infty} \frac{(n+1)^2}{(2n+2)(2n+1)}=\frac{1}{4}<1$
Therefore $\displaystyle \sum_{n\in\mathbb{N}}\frac{(n!)^2}{(2n)!}$ converges by the ratio test. Thus by definition $\displaystyle \frac{(n!)^2}{(2n)!}\to 0$