Is this series convergent or divergent $\sum_{n=1}^\infty{\frac{(2n)!}{n^{2n}}}$?

Question

I have problem with $$\sum_{n=1}^\infty{\dfrac{(2n)!}{n^{2n}}} $$ I try to use Cauchy Condensation Test, but unsuccessfully.

Any suggestions? Thanks for any help.

Answer 1

use ratio test $$L=\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow \infty }\dfrac{(2n+2)!}{(n+1)^{2n+2}}\dfrac{n^{2n}}{(2n)!}=\lim_{n\rightarrow \infty }\frac{(2n+2)(2n+1)}{(n+1)^2(1+\frac{1}{n})^{2n}}=\lim_{n\rightarrow \infty }\frac{(2+2/n)(2+1/n)}{(1+1/n)^2(1+\frac{1}{n})^{2n}}=\frac{4}{e^2}<1$$ so the series converges

Answer 2

Ratio Test (together with the known limit of $(1+1/n)^n$), or Stirling's approximation to the factorial.

Answer 3

By AM-GM you have for all $1 \leq k \leq n-1$:

$$k \cdot (2n-k) \leq n^2 $$

Using this for $ 3 \leq k \leq n-1$ you get

$$\dfrac{(2n)!}{n^{2n}} = \frac{1 \cdot (2n-1) \cdot 2 \cdot (2n-2) \cdot 2n}{n^5} \cdot \left(\prod_{k=3}^{n-1} \frac{k(2n-k)}{n^2} \right)\cdot \frac{n}{n} \\ \leq\frac{4 \cdot (2n-1) \cdot (2n-2)}{n^4}$$

As $\sum \frac{4 \cdot (2n-1) \cdot (2n-2)}{n^4}$ is convergent.....

Answer 4

Well, I think You should use D'Alembert's criterion, because You have a factorial in this series.

$$\lim_{n\to\infty}{\dfrac{(2(n+1))! n^{2n}}{n^{2(n+1)}(2n)!}}$$ $$\lim_{n\to\infty}{\dfrac{(2n)!(2n+2)(2n+1)n^{2n}}{n^{2n}n^2(2n)!}}$$ $$\lim_{n\to\infty}{\dfrac{4n^2+6n+2}{n^2}=4>1}$$

If $\lim_{n\to\infty}{\dfrac{a_{n+1}}{a_n}}>1$ this series is divergent.

So, $\sum_{n=1}^\infty{\dfrac{(2n)!}{n^{2n}}}$ is divergent.