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?

To carry out the integration of
$y+1=(1-2x)\frac{dy}{dx}
$
we have
$\dfrac1{1-2x}
=\dfrac{y+1}{(y+1)'}
=(\ln(y+1))'
$
so
$\ln(y+1)
=\int \dfrac{dx}{1-2x}
=-\frac12\ln(1-2x)+c
$
so
$y+1 
=\dfrac{C}{\sqrt{1-2x}}
$
so
$y
=\dfrac{C}{\sqrt{1-2x}}-1
$.
At $x=0, y=0$ so
$C=1$.
Then,
for $x = \frac25$,
$y 
= \dfrac{1}{\sqrt{1/5}}-1
=\sqrt{5}-1
$.
