<s>[INST] Evaluate $\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdot \ldots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$ 
Evaluate $$\sum_{r=1}^{\infty} \frac{1 \cdot 3  \cdots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$$

Let $$y=x + \frac{1 \cdot 3 \cdot}{2!} x^2 + \frac{1 \cdot 3 \cdot 5}{3!} x^3+\ldots$$ be the given expression.(replacing $2/5$ with $x$)
After some manipulations,
$$y+1=(1-2x)\frac{dy}{dx}$$
Integrating and substituting $x=\dfrac{2}{5}$, we get $y=\sqrt{5}-1$.
Is there any other way to solve this question?
 [/INST] $$S=\sum_{r=1}^{\infty} \frac{1.3.5....(2r-1)}{r!}\left(\frac{2}{5}\right)^r$$
$$\implies S=\sum_{r=1}^{\infty} \frac{(2r)!}{r!~ r!} 5^{-r} =\sum_{r=1}^{\infty} {2r \choose r} 5^{-r}=\frac{1}{\sqrt{1-4/5}}-1=\sqrt{5}-1.$$
Here we have used $$\sum_{k=0}^{\infty} {2k \choose k} x^k=\frac{1}{\sqrt{1-4x}}$$
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