Determine convergence $\sum_{n=0}^{\infty}\frac{n!\cdot 2^n}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n+1)} $

Question

I tried Ratio test for this series and got 1.

$$\sum_{n=0}^{\infty}\frac{n!\cdot 2^n}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n+1)}$$

Any suggestions? I believe the limit of the sequence goes to $\infty$ , but I don't know how to prove.

Thanks a lot!

Answer 1

Since \begin{align}\frac{n!2^n}{1\cdot3\cdot5\cdots(2n+1)}=&\frac{2\cdot4\cdot6\cdots2n}{1\cdot3\cdot5\cdots(2n+1)}\\=&2\cdot\frac{4}{3}\cdot\frac{6}{5}\cdots\frac{2n}{2n-1}\cdot\frac{1}{2n+1}\\ \geq&\frac{2}{2n+1}\geq\frac{2}{2n+2}=\frac{1}{n+1}\end{align} we have that $$ \sum_{n=0}^\infty\frac{n!2^n}{1\cdot3\cdot5\cdots(2n+1)}\geq\sum_{n=0}^\infty\frac{1}{n+1}=\infty. $$

Answer 2

Hint. One may observe that, as $n \to \infty$, we have $$ \frac{n!\cdot 2^n}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n+1)}=\frac{(n!)^2\cdot 2^{2n}}{(2n+1)!} \sim \frac{\sqrt{\pi}}2\cdot \frac1{\sqrt{n}} $$ by using Stirling's approximation formula, yielding the divergence of the initial series by the comparison test.

Answer 3

The general term is $$\frac{n!2^n}{(2n+1)!!}=\frac{(2n)!!}{(2n+1)!!}=\prod_{k=1}^{n}\left(1-\frac{1}{2k+1}\right)=\frac{1}{2}\,B\left(n+1,\frac{1}{2}\right)$$ that behaves like $\frac{1}{2}\sqrt{\frac{\pi}{n}}$ by Gautschi's inequality. It follows that the given series is divergent by the $p$-test. An alternative approach is to notice that $$ \sum_{n\geq 0}\frac{n! (2x)^n}{(2n+1)!!} = \frac{\arcsin(\sqrt{x})}{\sqrt{x(1-x)}} $$ for any $x\in(0,1)$, but the RHS has a singularity at $x=1$.
With a more elementary approach we may notice that: $$ \sum_{n=0}^{M}\frac{n!2^n}{(2n+1)!!}=-1+\frac{(2M+2)!!}{(2M+1)!!} $$ by creative telescoping, and the conclusion is just the same.