Does this series with binomial coefficient converge (absolutely)? How do I find out if this series converges (absolutely)?
$$\sum_{k=0}^\infty \binom {2k}k * 5 ^{-k} $$
We have the binomial coefficient there and I was wondering if one can actually cancel the k! in the numerator  ?

Regarding the denominator, it 
$$5^{-k} = (1/5^k)$$  converges as it is getting bigger and bigger. 
 A: Note that the series with positive terms therefore it doesn't make sense talk about absolute convergence.
We can use that by Stirling's approximation
$$\binom {2k}k\sim \frac{4^k}{\sqrt{\pi k}}$$
therefore
$$\binom {2k}k \cdot 5 ^{-k}\sim \frac{4^k}{5^k\sqrt{\pi k}}$$
from which we can conclude by limit comparison test.
As a simpler alternative by ratio test
$$\frac{\binom {2k+2}{k+1} \frac1{5 ^{k+1}}}{\binom {2k}{k} \frac1{5 ^{k}}}=\frac15 \frac{(2k+2)!(k!)^2}{[(k+1)!]^2(2k)!}=\frac15 \frac{(2k+2)(2k+1)}{(k+1)^2}\to \frac45$$
A: The generating function for the central binomial coefficients is well known :
$$\tag{1}\frac 1{\sqrt{1-4x}}=\sum_{k=0}^\infty \binom {2k}k\, x^k$$
(you may prove it using the Taylor series for $\,(1-4x)^{-1/2}\;$)
as well as the asymptotic expression for the binomial coefficients :
(for example using Stirling as done by gimusi)
$$\tag{2}{2k \choose k} \sim \frac{4^k}{\sqrt{\pi k}}$$
so that $(1)$ will converge for $\;x< \dfrac 14$.
For $x=\dfrac 15$ the series becomes $\;\displaystyle\frac 1{\sqrt{1-\frac 45}}=\sqrt{5}$
