How to bound $\binom{2n-1}{n}$

Question

I am trying to prove that the following series is convergent :

$$\sum_{n=1}^{\infty} \frac{(2n-1)!}{(2^{n} n!)^{2}}$$

If the numerator is $(2n)!$ then it is easily done by CT with $\frac{1}{2n+1}$; however for this case I can't find out how to bound it. (LCT does not work since the limit converges to $1$.)

I have found an upper bound for the sequence but it is no use since it is similar with $\frac{1}{n}$ which diverges. Can anyone help?

EDIT : I think the following question would be an answer since $\frac{(2n-1)!}{(2^{n} n!)^{2}} \le \frac{1}{2n\sqrt{3n+1}}$ and the $p$ series test.

Link

Just notice that the sum is smaller than the case when the numerator is $\left(2n\right)!$ and then complete the proof with what you said, also what you are summing is not exactly $2^{-2n}\binom{2n-1}{n}$, is $2^{-2n}\binom{2n}{n}\cdot\frac{1}{2n}$ — Raul Fernandes Horta, Dec 19 '22 at 01:43
You can use Stirling’s approximation or Raabe’s convergence test for example. — KStarGamer, Dec 19 '22 at 13:21

score 2 · Answer 1 · answered Dec 19 '22 at 06:20

This is $\displaystyle\sum_{n=1}^\infty\frac{a_n}{2n}$ with $\displaystyle a_n=\frac{(2n)!}{(2^n n!)^2}\asymp\frac1{\sqrt{n\pi}}$ (well-known; the weaker $a_n=O(n^{-1/2})$ is shown in a lot of posts here on MSE, including the one you mention, and even my own). Moreover, it's easy to evaluate the given sum in closed form, using $\sum_{n=0}^\infty a_n z^n=(1-z)^{-1/2}$: $$\sum_{n=1}^\infty\frac{a_n}{2n}=\int_0^1\left(\frac1{\sqrt{1-x^2}}-1\right)\frac{dx}{x}=-\log\left(1+\sqrt{1-x^2}\right)\Bigg|_0^1=\log2.$$

How to bound $\binom{2n-1}{n}$

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