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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!

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  • $\begingroup$ Hint: The series is divergent. $\endgroup$ – Cornman Jul 20 '18 at 18:07
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    $\begingroup$ @Comman - the OP declares he wants to prove that the series is divergent. How does "the series is divergent" provide a hint? $\endgroup$ – uniquesolution Jul 20 '18 at 18:08
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    $\begingroup$ Did you try Stirling's formula? $\endgroup$ – uniquesolution Jul 20 '18 at 18:08
  • $\begingroup$ @Cornman: how is that a hint in this context? Care to elaborate? $\endgroup$ – Clayton Jul 20 '18 at 18:10
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    $\begingroup$ @uniquesolution I do not know what you are reading, but I read "So I'm starting to suspect that the series might actually be convergent" $\endgroup$ – Cornman Jul 20 '18 at 18:10
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$$\frac{(2n)!}{4^n(n!)^2}=\frac{C_{2n}^n}{4^n}=\frac{\sum (C_n^i)^2}{4^n}\geq\frac{\frac{(\sum C_n^i)^2}{n+1}}{4^n}=\frac{1}{n+1}$$ by vandermonde identity and QM-AM inequality. and harmonic series is divergent so the series in question is also divergent.

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  • $\begingroup$ Can you expand a little on how we got the last three inequalities/equalities? $\endgroup$ – Shadowfirex Jul 20 '18 at 19:12
  • $\begingroup$ Sorry I don't know vandermonde identity and QM-AM, I'm reading about them right now. But any more details would be helpful too! $\endgroup$ – Shadowfirex Jul 20 '18 at 19:13
  • $\begingroup$ Makes sense thanks! $\endgroup$ – Shadowfirex Jul 22 '18 at 22:57
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Consider $$S(x)= \sum_{n=0}^\infty \binom{2n}{n} \left(\frac{x} {2} \right) ^{2 n} $$ Since $$\binom{\frac{-1}{2}}{n}=(-1)^n\frac{\binom{2n}{n}}{4^n}$$ By the binomial series we have that $$\sum_{n=0}^{\infty}\binom{\frac{-1}{2}}{n}(-x^2)^n=\frac{1}{\sqrt{1-x^2}}$$ thus your series is just $$S(1)=\frac{1}{\sqrt{1-1}}\rightarrow\infty$$

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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.

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  • $\begingroup$ $\frac{s_n}{2}\frac{2n+1}{n+1} \geq \frac{1}{n+1}\frac{2n+1}{2n+2} < \frac{1}{n+1}$ so $s_{n+1} \geq \frac{1}{n+1}$ is not true. $\endgroup$ – Shadowfirex Jul 20 '18 at 21:08
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    $\begingroup$ @Shadowfirex: Yeah, but you want $s_{n+1}\ge\frac1{n+2}$. $\endgroup$ – Vincenzo Oliva Jul 20 '18 at 21:20
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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!

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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.

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