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$$\sum_{n=1}^\infty {{4^n n!n!}\over{(2n)!}}$$

I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn't make any errors when doing the ratio test, it may diverge, but I need help proving that. Is there any other test I could try.

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Note that $$\frac{(2n)!}{n!n!}\leq\sum_0^{2n}\binom{2n}{k}=(1+1)^{2n}=4^n$$ So $$\frac{4^n n!n!}{(2n)!}\geq 1$$

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  • $\begingroup$ Do you think the series comparison test would help ? $\endgroup$ – user287927 Nov 7 '15 at 21:31
  • $\begingroup$ It would, but it's not required. The terms of your series don't tend to $0$, so the series cannot be convergent. $\endgroup$ – Wojowu Nov 7 '15 at 21:32
  • $\begingroup$ how would you show that it diverges $\endgroup$ – user287927 Nov 7 '15 at 21:35
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One approach is a comparison test, approximating the function using Stirling's approximation. Since $n!= \sqrt{2\pi}n^{n+1/2}\text{e}^{-n}(1+\mathcal{O}(\tfrac{1}{n}))$ we have $$\frac{4^n n!^2}{(2n)!}\approx\frac{4^{n}2\pi \cdot n^{2n+1}\text{e}^{-2n}}{\sqrt{2\pi} 2^{2n+1/2} n^{2n+1/2}\text{e}^{-2n}}=\sqrt{\pi n}.$$ The series therefore diverges.

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