<s>[INST] Closed form expression for infinite series I was given the following function: 
$$ f(x) =  x + \frac{2x^3}{1\cdot3} + \frac{2\cdot4x^5}{1\cdot3\cdot5} + \frac{2\cdot4\cdot6x^7}{1\cdot3\cdot5\cdot7}...   $$ $$ \forall x    \in [0,1) $$
And then I was asked to find the value of $ f(\frac{1}{\sqrt{2}}) $, which obviously requires me to compute the closed form expression of the infinite series.
I tried 'Integration as a limit of sum' but I was unable to modify the expression accordingly. How do I approach the problem?
 [/INST] I would suggest the following representation
$$\sum_{n=0}^{\infty}\frac{x^{2n+1}(2n)!!}{(2n+1)!!}=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}$$
Plug in $x=\frac{1}{\sqrt{2}}$. The result will be
$$\frac{\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{1-\frac{1}{2}}}=\frac{\sqrt{2}\pi}{4}$$
Update:
Your question is similar to problem $6(ii)$ in the eighth William Lowell Putnam Competition. I will present the solution from the book.

Let
  $$f(x) =  x + \frac{2x^3}{1\cdot3} + \frac{2\cdot4x^5}{1\cdot3\cdot5} + \frac{2\cdot4\cdot6x^7}{1\cdot3\cdot5\cdot7}\cdots   $$
  Then 
  $$\begin{align}
f'(x) &= 1+x\left[2x+\frac{2}{3}\cdot 4x^3+\frac{2}{3}\cdot\frac{4}{5}\cdot6x^5 +\cdots \right]\\
\\&= 1+x\frac{d}{dx}\left[x^2+\frac{2}{3}x^4+\frac{2}{3}\cdot\frac{4}{5}x^6+\cdots\right]\\
\\&=1+x\frac{d}{dx}(xf(x))=1+xf(x)+x^2f'(x).
\end{align}$$
  Thus $f'(x)$ satisfies the differential equation
  $$(1-x^2)f'(x)=1+xf(x)\tag{1}$$
  and the initial condition
  $$f(0)=0.\tag{2}$$
  (We note that the series for $f$ is convergent for $|x| < 1$ so that all formal manipulations are justified.)
  Now $(1)$ is a first order linear non-singular differential equation on the interval $(-1,1)$, so it has a unique solution satisfying (2).

The next step is to solve the differential equation, 
$$(1-x^2)\frac{df(x)}{dx}=1+xf(x)$$
Rewrite the equation:
$$\frac{df(x)}{dx}-\frac{xf(x)}{1-x^2}=\frac{1}{1-x^2}$$
Multiply both sides by $\sqrt{1-x^2}$
$$\frac{df(x)}{dx}\sqrt{1-x^2}-\frac{xf(x)}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$$
Substitute $\frac{x}{\sqrt{1-x^2}}=-\frac{d}{dx}\sqrt{1-x^2}$
$$\frac{df(x)}{dx}\sqrt{1-x^2}+\frac{d}{dx}\left(\sqrt{1-x^2}\right)f(x)=\frac{1}{\sqrt{1-x^2}}$$
Apply the reverse product rule $f\frac{dg}{dx}+g\frac{df}{dx}=\frac{d}{dx}(fg)$ to the left-hand side
$$\frac{d}{dx}\left(f(x)\sqrt{1-x^2}\right)= \frac{1}{\sqrt{x^2-1}}$$
Integrate both sides with respect to $x$
$$\int\frac{d}{dx}\left(f(x)\sqrt{1-x^2}\right)dx= \int\frac{1}{\sqrt{1-x^2}}dx$$
Recall fundamental theorem of calculus $\frac{d}{dx}\int f(x) dx =f(x)$ and
$\int\frac{1}{\sqrt{1-x^2}}=\arcsin(x)+C$, then
$$f(x)\sqrt{1-x^2}=\arcsin(x)+C$$
We have the initial condition $f(0)=0$, so
$$0=\arcsin(0)+C$$
$$C=0$$
Therefore,

$$f(x)=\frac{\arcsin(x)}{\sqrt{1-x^2}}$$

</s>