1. ## convergence problem

$\sum \frac{2^{n-1}n!}{n^{2n+1}}$
so, i should use D'Alambert rule. Then i got
$\frac{a_{n+1} }{a_{n}}= \frac{\frac{2^n(n+1)!}{(n+1)^{2n+3}}}{\frac{2^{n-1}n!}{n^{2n+1}}}$

and what now?

2. ## Re: convergence problem

Acknowledging that

$\frac{(n+1)!}{n!} = n+1$

and

$\frac{2^{n}}{2^{n-1}} = 2$

Your ratio can be simplified to

$\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$

$= \lim_{n \to \infty} \frac{2(n+1)\cdot n^{2n+1}}{(n+1)^{2n+3}}$

$= \lim_{n \to \infty} \frac{2(n+1)\cdot n^{2n+1}}{(n+1)^{2n+1}\cdot (n+1)^2}$

$= \lim_{n \to \infty} \left (\frac{n}{n+1}\right )^{2n+1} \cdot \frac{2}{n+1}$

$= \lim_{n \to \infty} \frac{2}{\left (1+\frac{1}{n}\right )^{2n+1} \cdot (n+1)}$

$= \lim_{n \to \infty} \frac{1}{\left (1+\frac{1}{n}\right )^{2n}} \cdot \lim_{n \to \infty} \frac{1}{1+\frac{1}{n}} \cdot \lim_{n \to \infty} \frac{2}{n+1}$

$= \frac{1}{e^2} \cdot 1 \cdot 0$

$= 0$

So the series converges.

3. ## Re: convergence problem

Originally Posted by DonnieDarko
$\sum \frac{2^{n-1}n!}{n^{2n+1}}$
so, i should use D'Alambert rule. Then i got
$\frac{a_{n+1} }{a_{n}}= \frac{\frac{2^n(n+1)!}{(n+1)^{2n+3}}}{\frac{2^{n-1}n!}{n^{2n+1}}}$

and what now?
Simplify.

$= \left( \frac{2^n}{2^{n-1}} \right) \left( \frac{(n+1)!}{n!} \right) \left( \frac{n^{2n+1}}{(n+1)^{2n+3}} \right) = 2(n+1) \left( \left( \frac{n}{(n+1)^3} \right) \left( \frac{n}{n+1} \right)^{2n} \right)$

$= \frac{2n}{(n+1)^2} \left( \frac{n}{n+1} \right)^{2n} = \frac{2n}{(n+1)^2} \left( \frac{n+1-1}{n+1} \right)^{2n} = \frac{2n}{(n+1)^2} \left( 1 - \frac{1}{n+1} \right) ^{2n}$

$= \frac{2n}{(n+1)^2} \left( \left( 1 - \frac{1}{n+1} \right) ^{n+1} \right) ^ {\frac{2n}{n+1}}$.

But $\left( 1 - \frac{1}{n+1} \right)^{n+1} \rightarrow e^{-1}, \frac{2n}{n+1} \rightarrow 2$, and $\frac{2n}{(n+1)^2} \rightarrow 0$, so

$\ \frac{a_{n+1}}{a_n} = \frac{2n}{(n+1)^2} \left( \left( 1 - \frac{1}{n+1} \right) ^{n+1} \right) ^ {\frac{2n}{n+1}} \rightarrow (0)(e^{-1})^2 = 0$ as $\ n \rightarrow \infty$.

4. ## Re: convergence problem

johnsomeone, your post contains an error. The term with the $2n+3$ exponent should be in the denominator.

5. ## Re: convergence problem

Thanks. It's hard to "see" algebra in LaTex!

6. ## Re: convergence problem

Thanks you guys a lot