Problem trying to form 1-a/n for Raabe's test in series convergence

Question

I have the following series: $$\sum_{k=1}^\infty \frac{(2k)!}{4^k(k!)^2}$$ Using ratio test I find: $\lim_{k\to \infty}\frac{a_{n+1}}{a_n}=\lim_{k\to \infty}\frac{2k+1}{2(k+1)}=1$ I'm trying to form $1-\frac{a}{k}$ since ratio test doesn't help.

$\frac{2k+1}{2(k+1)}=\frac{2k+2-1}{2k+2}=1-\frac{1}{2k+2}=1-\frac{1}{2k}\frac{1}{1+1/k}=1-\frac{1}{2k}(1+1/k)^{-1}$ With binomial expansion we have : $1-\frac{1}{2k}(1-1/k+1/k^2 -1/k^3+...)$ How do I proceed now? I think our teacher ignored the whole parenthesis somehow and said this : $1-\frac{1}{2k}(1-1/k+1/k^2 -1/k^3+...)\sim1-\frac{1}{2k}$but it doesn't make sense to me.

Answer 1

Using Raabe's test, we have

$$\begin{align} k\left(\frac{a_k}{a_{k+1}}-1\right)&=k\left(\frac{4(2k)!((k+1)!)^2}{(k!)^2(2k+2)!}-1\right)\\\\ &=k\left(\frac{2(k+1)}{2k+1}-1\right)\\\\ &=\frac{k}{2k+1}\to \frac12\,\,\,\text{as}\,\,k\to \infty \end{align}$$

Inasmuch as the limit is less than $1$, the tests guarantees that the series diverges.

Answer 2

It is a diverging series since $$ \frac{1}{4^k}\binom{2k}{k}\approx\frac{1}{\sqrt{\pi k}}.\tag{1} $$

You may notice that the square of the LHS is given by

$$\prod_{j=1}^{k}\left(1-\frac{1}{2j}\right)^2 = \frac{1}{4}\prod_{j=2}^{k}\left(1-\frac{1}{j}\right)\prod_{j=2}^{k}\left(1-\frac{1}{(2j-1)^2}\right)^{-1} \\=\frac{1}{4k}\prod_{j=2}^{k}\left(1-\frac{1}{(2j-1)^2}\right)^{-1}\tag{2}$$ and the infinite product $\prod_{j\geq 2}\left(1-\frac{1}{(2j-1)^2}\right)^{-1}$ is convergent to $\frac{4}{\pi}$.