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Does the series

$$\sum_{n = 1} ^ {\infty} \frac {(2n)!}{2^{2n}(n!)^2}$$

converge? The ratio test doesn't work for the series.

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  • $\begingroup$ Try comparison test. $\endgroup$ – evil999man May 18 '14 at 17:35
  • $\begingroup$ As an aside, $\quad\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}n^2}=2\arcsin^2x$, where $\displaystyle{2n\choose n}=\frac{(2n)!}{\big(n!\big)^2}$ $\endgroup$ – Lucian May 18 '14 at 18:52
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If you want to avoid Stirling's formula, note that $${2n\choose n}{1\over 2^{2n}}={1\over 2n}\prod_{j=1}^{n-1}\left(1+{1\over 2j}\right)\geq {1\over 2n},$$ so your series diverges by comparison with the harmonic series.

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From Stirling's formula

$$n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^{n}$$

we have

$$\frac {(2n)!}{2^{2n}(n!)^2}\sim \frac {\sqrt{4 \pi n}}{2^{2n}(n!)^2} \left(\frac{2n}{e}\right)^{2n}=\frac {\sqrt{4 \pi n}}{(n!)^2} \left(\frac{n}{e}\right)^{2n}$$

and applying again Stirling's formula to denominator:

$$\frac {\sqrt{4 \pi n}}{(n!)^2} \left(\frac{n}{e}\right)^{2n}\sim \frac {\sqrt{4 \pi n}}{2\pi n} \left(\frac{e}{n}\right)^{2n} \left(\frac{n}{e}\right)^{2n}=\frac{\sqrt{4 \pi n}}{2\pi n}\sim \frac{1}{\sqrt n}$$

But

$$\sum_n \frac{1}{\sqrt n}$$

is an harmonic divergent series.

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