Question about a "trick" to solve a limit of a sequence ..

**TWiX** · November 10th 2009, 01:06 PM

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 ?

**Drexel28** · November 10th 2009, 01:56 PM

Originally Posted by TWiX

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 ?

I am not entirely sure what the trick you are refering to is. It is true that if $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|< 1$ that $\lim_{n\to\infty}a_n=0$ . Is that what you are asking?

**TWiX** · November 11th 2009, 04:30 AM

Yeah ..
Actually if the series Sigma(an) is convergent then the limit of an = 0

**Drexel28** · November 11th 2009, 08:25 AM

Originally Posted by TWiX

Yeah ..
Actually if the series Sigma(an) is convergent then the limit of an = 0

Yes that is true.

So for any spectators what is going on is that TWiX is just making the observation that if $\sum_{n\in\mathbb{N}}\sigma_n$ converges it is necessary for $\lim_{n\to\infty}\sigma_n$ . And of course there are a lot of ways to prove that $\sum_{n\in\mathbb{N}}\sigma_n$ converges thaty may be easier than actually computing $\lim_{n\to\infty}\sigma_n$ .

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

Solution: Note that $\lim_{n\to\infty}\left|\frac{\left[(n+1)!\right]^2}{(2(n+1))!}\cdot\frac{(2n)!}{(n!)^2}\right|$ $=\lim_{n\to\infty}\frac{(n+1)^2(n!)^2}{(2n+2)(2n+1 )(2n)!}\cdot\frac{(2n)!}{(n!)^2}=\lim_{n\to\infty} \frac{(n+1)^2}{(2n+2)(2n+1)}=\frac{1}{4}<1$

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

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