I tried using the ratio test but it fails. This can be solved using Raabe test but is there any other way of solving this ? One argument was that numerator will always be smaller than denominator so the nth term goes to zero. But this can't be used to prove convergence. Any help is appreciated!
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I suggest rewriting the terms using factorials, then invoke Stirling's aproximation. – lulu Apr 12 '22 at 15:10
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As pointed out by @lulu - you can rewrite as $\sum_{n=1}^\infty \frac{(2n)!}{2^{2n} (n!)^2}$ – Gregory Apr 12 '22 at 15:25
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1Just a question for @Hãru2you: What is wrong with Raabe tests? As mentioned in another comment, one can invoke Stirling's approximation to factorials which indeed reduces the problem considerable, but proving the validity of the approximation (Stirling's approximation theorem, that is) requires much more effort that Rabee's etst. – Mittens Apr 12 '22 at 15:27
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In general the numerator always being smaller than the denominator does not force the $n$th term to zero: consider $\frac{n+1}{2n}$ as an example – Henry Apr 12 '22 at 15:35
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1Do you know bounds on $\binom{2n}{n}$ like the ones here? – 2'5 9'2 Apr 12 '22 at 16:21
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Hint: try using simple algebra (no need for Stirling's approximation) to prove by mathematical induction that the $n^\text{th}$ term is bounded below by a constant multiple of $1/\sqrt{n}.$ – Calum Gilhooley Apr 12 '22 at 16:51
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@alex.jordan Useful reference - thanks. I see now that my comment was a bit redundant. I just happened to be computing similar bounds a couple of weeks ago, inspired by this Maths.SE answer. – Calum Gilhooley Apr 12 '22 at 17:20