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.