Show that $\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}$ is divergent

Question

Given the series $S=\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}$ I'm trying to prove its divergent but with no luck. It doesn't tend to $\infty$ as $n$ grows large and ratio test is inconclusive. I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?

Thanks!

Answer 1

$$\frac{(2n)!}{4^n(n!)^2}=\frac{C_{2n}^n}{4^n}=\frac{\sum (C_n^i)^2}{4^n}\geq\frac{\frac{(\sum C_n^i)^2}{n+1}}{4^n}=\frac{1}{n+1}$$ by vandermonde identity and QM-AM inequality. and harmonic series is divergent so the series in question is also divergent.

Answer 2

Use Raabe's Test with $$\dfrac{a_{n+1}}{a_{n}}=1-\dfrac{A}{n}+\dfrac{A_n}{n}=1-\dfrac{\frac12}{n}+\dfrac{A_n}{n}$$ where $A_n=\dfrac{1}{2(n+1)}\to0$ as $n\to\infty$, then $A=\dfrac12<1$ shows the series diverges!

Answer 3

Since in a comment you said you're not familiar with QM-AM and Vandermonde's identity, you may find the following useful.

Hint: Note that$$s_n=\frac{(2n)!}{4^n(n!)^2}=\frac1{4^n}\frac{(n+1)(n+2)\cdots(2n)}{n!}$$so one has $$s_{n+1}=s_n\frac{(2n+1)(2n+2)}{4(n+1)^2}=\frac{s_n}{2}\frac{2n+1}{n+1}.$$Use this to prove $s_n\ge\frac1{n+1}$ by induction.

Answer 4

Whatare you allowed to use? With Stirling's approximation $$ \binom{2n}{n} \sim \frac{4^n}{\sqrt{ \pi n}} $$ so the summand is $O(\frac{1}{\sqrt{n}})$ and

$$ \sum_{k=1}^{n} \frac{1}{\sqrt{\pi k}} $$ and this diverges by comparing it to the Harmonic series.

Once again, this works if you are allowed to use the divergence of Harmonic series and Stirling's approximation.