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Consider the formal series $$\sum_{n=1}^\infty \frac{(n!)^2}{(2n)!}4^n.$$

Is there a way to test for the convergence/divergence of this series without explicitly using Stirling's approximation?

We have $$\lim_{n\to\infty} \frac{(n!)^2}{(2n)!}4^n=\lim_{n\to\infty}4^n\frac{n}{2n}\frac{n-1}{2n-1}\cdots\frac{2}{n+2}\frac{1}{n+1}.$$

What is the limit? And if the limit turned out to be 0, what can we do?


marked as duplicate by José Carlos Santos sequences-and-series Sep 26 at 9:37

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  • 1
    $\begingroup$ Try the ratio test. $\endgroup$ – Ninad Munshi Sep 26 at 9:34
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    $\begingroup$ @NinadMunshi Ratio test doesn't work. $\endgroup$ – Kabo Murphy Sep 26 at 9:34
  • $\begingroup$ @NinadMunshi ratio test fails. $\endgroup$ – UserA Sep 26 at 9:35