Evaluate $\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdot \ldots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$

$$S=\sum_{r=1}^{\infty} \frac{1.3.5....(2r-1)}{r!}\left(\frac{2}{5}\right)^r$$ $$\implies S=\sum_{r=1}^{\infty} \frac{(2r)!}{r!~ r!} 5^{-r} =\sum_{r=1}^{\infty} {2r \choose r} 5^{-r}=\frac{1}{\sqrt{1-4/5}}-1=\sqrt{5}-1.$$ Here we have used $$\sum_{k=0}^{\infty} {2k \choose k} x^k=\frac{1}{\sqrt{1-4x}}$$

The hint:

It's a Taylor expansion for $f(x)=\frac{1}{\sqrt{1-2x}}-1.$