<s>[INST] Convergence of the series $\sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} 2^n= \sum_{n=1}^\infty \dfrac{4^n n! n!}{(2n)!}$ I am trying to compute the interval of convergence of the power series $\sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} x^{2n+1}$. Letting $a_n= \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} x^{2n+1}$, we have $\lim_{n\to \infty} |a_{n+1}/a_n|= x^2/2$, so the radius of convergence is $\sqrt{2}$. What I only need to check now is that checking the convergence of the series above when $x=\pm \sqrt{2}$. Since $\sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} \sqrt{2}^{2n+1}=-\sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} (-\sqrt{2})^{2n+1}$, I only need to consider the case $x=\sqrt{2}$. In the case $x=\sqrt{2}$, the series is given by $\sqrt{2} \sum_{n=1}^\infty \dfrac{n!}{1\cdot 3 \cdot \cdots \cdot (2n-1)} 2^n= \sqrt{2} \sum_{n=1}^\infty \dfrac{4^n n! n!}{(2n)!}$, but I can't see whether or not this series converges. Any hints?
 [/INST] One more, possibly easier, way to use estimation:
$$\sqrt{2n} \lt  \frac{2}{1}\cdot\frac{4}{3}\cdot \frac{6}{5}\cdots \frac{2n}{2n-1} \leqslant 2\sqrt{n}$$
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