# Does $\sum_1^\infty\frac{(n!)^2+(2n)^n}{n^{2n}}$ converge? (without Stirling's approximation)

This question have been asked again a long ago here, but its only answer gives just a hint about Stirling's approximation.

I am trying to study the convergence $$\sum_1^\infty\frac{(n!)^2+(2n)^n}{n^{2n}}$$, but without Stirling's approximation.

I tried Cauchy's condensation test with no luck. Wolfram alpha suggests root test

The limit is:

$$\lim_{n\to\infty} \sqrt[n]{\frac{(n!)^2+(2n)^n}{n^{2n}}}$$

Again Wolfram calulates the limit to be: $$e^{-2}$$

Therefore, somehow $$\lim_{n\to\infty}\sqrt[n]{\frac{(n!)^2+(2n)^n}{n^{2n}}} = \lim_{n\to\infty} \left(1-\frac2{n}\right)^n$$

but I have no idea how to prove that.

I don't want to reduce this question exclusively to the calculation of this limit, any answer that doesn't include Stirling's approximation is welcome.

Thanks

• What Stirling implies is that the numerator is dominated by the $n!^2$ term. So the question boils down to $(n!)^{1/n}/n\to e^{-1}$ which I suppose is a weak form of Stirling. You might be able to get this by applying the trapezium rule to $\int_1^n\ln x\,dx$ and using crude estimates. – Angina Seng Aug 24 '20 at 11:51
• – Joe Aug 24 '20 at 11:55

Note that$$\require{cancel}\frac{\frac{(n+1)!^2}{(n+1)^{2(n+1)}}}{\frac{n!^2}{n^{2n}}}=\cancel{(n+1)^2}\left(\frac{n}{n+1}\right)^{2n}\frac1{\cancel{(n+1)^2}}\to\frac1{e^2}$$and that$$\frac{\frac{(2(n+1))^{n+1}}{(n+1)^{2(n+1)}}}{\frac{(2n)^n}{n^{2n}}}=2(n+1)\left(\frac{n+1}n\right)^n\left(\frac n{n+1}\right)^{2n}\frac1{(n+1)^2}=\frac2{n+1}\left(\frac n{n+1}\right)^n\to0.$$So, both series$$\sum_{n=1}^\infty\frac{n!^2}{n^{2n}}\text{ and }\sum_{n=1}^\infty\frac{(2n)^n}{n^{2n}}$$converge and therefore so does their sum.

• just finished calculating exactly same. +1. – zkutch Aug 24 '20 at 12:04
• Note that $\sum_1^\infty\frac{(n!)^2+(2n)^n}{n^{2n}} < \sum_1^\infty\frac{(n!)^2}{n^{2n}}$, and visa versa hence, calculating only one limit is sufficient. Thank you! – DimitrisPython Aug 24 '20 at 12:17
• How do you justify that inequality? – José Carlos Santos Aug 24 '20 at 12:20
• @JoséCarlosSantos I don't, its wrong. Thank you – DimitrisPython Aug 24 '20 at 12:36

Convergence of this series is quite easy. First note that $$\sum \frac {(2n)^{n}} {n^{2n}}$$ is convergent since it is dominated by $$\frac {2^{n}} {3^{n}}$$ if you omit the first two terms. Now use the fact that $$n! < (1)(2)(n^{n-2})$$ for $$n \geq 3$$. Can you finish?

We have that eventually, using this result

$$(n!)^2\ge \left(\frac{n^2}{e^2}\right)^n\ge(2n)^n$$

therefore

$$\frac{(n!)^2+(2n)^n}{n^{2n}}\le \frac{2(n!)^2}{n^{2n}}=2\frac{n!}{n^{n}}\frac{n!}{n^{n}} \le \frac2{n^2}$$

since

$$\frac{n!}{n^{n}}=\frac{1\cdot 2\cdots n}{n\cdot n\cdots n} \le \frac{1}{n}\cdot 1\cdot 1\cdots 1 = \frac{1}{n}$$