Determine if it converges or diverges : $\sum_{n=1}^{\infty} \frac {2^n \cdot n!}{1\cdot2\cdots (2n-1)}\cdot \frac{1}{\sqrt{2n+1}}$

Question

Here's the series:

$$\sum_{n=1}^\infty \frac {2^n \cdot n!}{1\cdot2\cdots (2n-1)}\cdot \frac{1}{\sqrt{2n+1}}$$

Does it converge or diverge ? Thanks

Answer 1

I guess the question has been answered, so some more thoughts:

• $\frac{n!}{(2n-1)!}=\left(2n\cdot\frac{n!}{2n!}\right)=2n\cdot\frac{1}{\prod_{k=1}^n(n+k)}<2n\cdot n^{-n}$

And note how this absorbs $2^n$ to make $2n\cdot \left(\frac{n}{2}\right)^{-n}$.

And btw. this is a really small quantity.

For the $\frac{1}{\sqrt{2n+1}}$, we have

• $2n\frac{1}{\sqrt{2n+1}}<\frac{2n}{\sqrt{2n}}=\sqrt{2n}$

Hence

$\frac {2^n \cdot n!}{1\cdot2\cdot...\cdot (2n-1)}\cdot \frac{1}{\sqrt{2n+1}}<\sqrt{2n}\,\left(\frac{n}{2}\right)^{-n}=c\cdot\left(\frac{n}{2}\right)^{1/2-n}$

So any presumably divergent remaining sum is smaller than some value of the zeta function on the real line, which are all small.

Answer 2

Using the ratio test I got the ratio is $$\lambda_n= \frac{(n+1)}{(2n+1)n} \sqrt{\frac{2n+1}{2n+3}}$$ obviously $$\displaystyle \lim_{n\to +\infty} \lambda_n=0$$ Therefore the sum should converge.