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?
$$S=\sum_{r=1}^{\infty} \frac{1.3.5....(2r-1)}{r!}\left(\frac{2}{5}\right)^r$$ $$\implies S=\sum_{r=1}^{\infty} \frac{(2r)!}{r!~ r!} 5^{-r} =\sum_{r=1}^{\infty} {2r \choose r} 5^{-r}=\frac{1}{\sqrt{1-4/5}}-1=\sqrt{5}-1.$$ 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 $f(x)=\frac{1}{\sqrt{1-2x}}-1.$
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 $.
