# 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?

The hint:

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

• @Qurultay I got it by the Taylor expansion. By the way, we obtain $\frac{1}{\sqrt{1-2x}}-1$ ;) – Michael Rozenberg Mar 15 '20 at 11:34

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

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 $$x = \frac25$$, $$y = \dfrac{1}{\sqrt{1/5}}-1 =\sqrt{5}-1$$.