As stated by the prompt, I'm looking to show the convergence of the series $$\sum_{n=0}^{\infty}\frac{1}{2}\cdot \frac{(2n)!}{n!(n+1)!}\cdot \left(\frac{1}{4}\right)^n$$ I've tried using the ratio, root, and Limit Comparison test (w/ a geometric series), and all have given me inconclusive answers. If anyone has some insights for how I might be able to show this series converges, your help would be much appreciated.
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1You can simplify your exptession or it is $(2n)!. – hamam_Abdallah Nov 25 '20 at 21:45
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Have you tried Stirling Formula. – hamam_Abdallah Nov 25 '20 at 21:47
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@hamam_Abdallah it is $(2n)!$, and I hadn't thought to use Stirling's approximation for the factorial, that might be a good route for me to explore. – Shaun Nov 25 '20 at 21:56
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One possible solution using Raabe's Test $$n\cdot\left(\frac{a_n}{a_{n+1}} -1\right) = n\cdot\frac{6n +6}{(2n+1)(2n+2)}\to \frac{3}{2}>1$$
zkutch
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That will do it for me, thank you! @J.G.'s answer was great and slick, but I think this will work better given the audience this proof would be written for. Thank you again – Shaun Nov 27 '20 at 08:50
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In fact, if $|x|\le\tfrac14$ then $\sum_{n\ge0}\frac{(2n)!}{n!(n+1)!}x^n=\frac{1-\sqrt{1-4x}}{2x}$, so the given sum is $1$.
J.G.
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J.G. Thank you for your quick response, I appreciate it. Your reference is a very straightforward approach, though I'm also curious if there might be other ways to show the series converges? – Shaun Nov 25 '20 at 21:58
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Answer doesn't look right. As $x\to 0$, series $\to 1$, while function $\to \infty$. – herb steinberg Nov 25 '20 at 22:01
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@herbsteinberg You are mistaken:$$\lim_{x\to0}\frac{1-\sqrt{1-4x}}{2x}=\lim_{x\to0}\frac{2}{1+\sqrt{1-4x}}=1.$$ – J.G. Nov 25 '20 at 22:03
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