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 ?

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- Nov 10th 2009, 01:06 PMTWiXQuestion about a "trick" to solve a limit of a sequence ..
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 ? - Nov 10th 2009, 01:56 PMDrexel28
- Nov 11th 2009, 04:30 AMTWiX
Yeah ..

Actually if the series Sigma(an) is convergent then the limit of an = 0 - Nov 11th 2009, 08:25 AMDrexel28
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 $\displaystyle \lim_{n\to\infty}\sigma_n$.

Compute $\displaystyle \lim_{n\to\infty}\frac{(n!)^2}{(2n)!}$**Example:**

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$__Solution:__

Therefore $\displaystyle \sum_{n\in\mathbb{N}}\frac{(n!)^2}{(2n)!}$ converges by the ratio test. Thus by definition $\displaystyle \frac{(n!)^2}{(2n)!}\to 0$