Convergence tests

[canvas]

Does the series converge or diverge?

[math]\sum_{n=1}^\infty\frac{(2n)!}{n^{2n}}\\\\\text{I tried to use d'alembert test but I got so hard limit, so I thought about cauchy's test:}\\\\\lim_{n\to\infty}\sqrt[n]{\frac{(2n)!}{n^{2n}}}\\\\\text{Do u have any ideas how to calculate this limit?}[/math]

 

[skeeter]

Try the ratio test?

 

[canvas]

Try the ratio test?

d'alembert test also means ratio test, just look what a horrible limit we'll get using this test... :/

 

[BigBeachBanana]

d'alembert test also means ratio test, just look what a horrible limit we'll get using this test... :/

Ratio tests are typically used for series with factorials because they cancel out nicely.

 

[blamocur]

The ratio test seems to work for me.

 

[canvas]

[math]\text{let}\,\,\,a_n=\frac{(2n)!}{n^{2n}}\\\\\Rightarrow a_{n+1}=\frac{[2(n+1)]!}{(n+1)^{2(n+1)}}=\frac{(2n+2)!}{(n+1)^{2n+2}}=\frac{(2n)!(2n+1)(2n+2)}{(n+1)^{2n}\cdot(n+1)^2}\\\\\text{so}\,\,\,\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(2n)!(2n+1)(2n+2)}{(n+1)^{2n}\cdot(n+1)^2}\cdot\frac{n^{2n}}{(2n)!}=\\\\=\lim_{n\to\infty}\frac{2(2n+1)n^{2n}}{(n+1)^{2n}(n+1)}=...\\\\\text{and here is the point where I stuck, any ideas?}[/math]

 

[skeeter]

[imath]\displaystyle \lim_{n \to \infty} \dfrac{(2n+2)(2n+1)}{(n+1)^2} \div \left(\dfrac{n+1}{n}\right)^{2n}[/imath]

try again …

 

[skeeter]

\(\displaystyle \frac{(2n + 2)!}{n^{2n + 2}} \cdot \frac{n^{2n}}{(2n)!} = \frac{(2n+2)(2n+1)(2n)!}{n^{2n}n^2} \cdot \frac{n^{2n}}{(2n)!}\)

Simplify and use \(\displaystyle \lim_{n\to\infty}\)

(n+1)st term of [imath] n^{2n} \text{ is } (n+1)^{2(n+1)} = (n+1)^{2n} \cdot (n+1)^2[/imath]

 

[nasi112]

\(\displaystyle \frac{(2n + 2)!}{(n + 1)^{2n + 2}} \cdot \frac{n^{2n}}{(2n)!} = \frac{(2n+2)(2n+1)(2n)!}{(n+1)^{2n}(n+1)^2} \cdot \frac{n^{2n}}{(2n)!}\)

Simplify and use \(\displaystyle \lim_{n\to\infty}\)

I think that you will need to use the logarithm technique to solve this limit. It is not difficult.

 

[nasi112]

(n+1)st term of [imath] n^{2n} \text{ is } (n+1)^{2(n+1)} = (n+1)^{2n} \cdot (n+1)^2[/imath]

Yeah skeeter. I spotted my mistake

 

[skeeter]

[imath]\displaystyle \lim_{n \to \infty} \dfrac{(2n+2)(2n+1)}{(n+1)^2} \div \left(\dfrac{n+1}{n}\right)^{2n}[/imath]

try again …

[imath]\left(\dfrac{n+1}{n}\right)^{2n} = \bigg[\left(1+\dfrac{1}{n}\right)^n\bigg]^2[/imath]

limit of the binomial inside the big [brackets] should be very familiar

 

[canvas]

Sorry guys I was completely blind, it was to late... so finally:

[math]\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{2(2n+1)}{n+1}:\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^{2n}=4:e^2=\frac{4}{e^2}<1\\\\\Rightarrow \sum_{n=1}^\infty\frac{(2n)!}{n^{2n}} - \text{converges!}[/math]