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?
 A: $${\frac {\arcsin \left( x \right) }{\sqrt {1-{x}^{2}}}}=x+{\frac {2}{
3}}{x}^{3}+{\frac {8}{15}}{x}^{5}+{\frac {16}{35}}{x}^{7}+O \left( {x}
^{9} \right) 
$$
Then
$$f\left(\frac{1}{\sqrt2}\right)
=\frac{\arcsin\left(\frac{1}{\sqrt2}\right)}{\sqrt{1-\frac12}}=\frac{\pi\sqrt2}{4}$$
A: 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,

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

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\left.\vphantom{\Large A}\mrm{f}\pars{x}\right\vert_{\ x\ \in\ \left[0,1\right)} \equiv  x + {2x^{3} \over 1 \times 3} + {2 \times 4x^{5} \over 1 \times 3 \times 5} +
{2 \times 4 \times 6x^{7} \over 1 \times 3 \times 5 \times 7}
+ \cdots:\ {\LARGE ?}}$.

$\underline{\large\texttt{An Explicit Evaluation:}}$
\begin{align}
\left.\vphantom{\Large A}\mrm{f}\pars{x}\right\vert_{\ x\ \in\ \left[0,1\right)} & \equiv
x + \sum_{n = 1}^{\infty}{\pars{\prod_{k = 1}^{n}2k}x^{2n + 1} \over \prod_{k = 1}^{n}\pars{2k + 1}} =
x + \sum_{n = 1}^{\infty}{\pars{2^{n}n!} \over
2^{n}\prod_{k = 1}^{n}\pars{k + 1/2}}\,x^{2n + 1}
\\[5mm] & =
x + \sum_{n = 1}^{\infty}{n! \over
\pars{3/2}^{\large\overline{n}}}\,x^{2n + 1} =
x + \sum_{n = 1}^{\infty}{n! \over
\Gamma\pars{3/2 + n}/\Gamma\pars{3/2}}\,x^{2n + 1}
\\[5mm] & =
x + \sum_{n = 1}^{\infty}n!\bracks{{1 \over \pars{n - 1}!}\,{\Gamma\pars{n}\Gamma\pars{3/2} \over
\Gamma\pars{n + 3/2}}}x^{2n + 1}
\\[5mm] & =
x + \sum_{n = 1}^{\infty}n
\bracks{\int_{0}^{1}t^{n - 1}\pars{1 - t}^{1/2}\,\dd t}x^{2n + 1}
\\[5mm] & =
x + x\int_{0}^{1}{\pars{1 - t}^{1/2} \over t}\
\underbrace{\sum_{n = 1}^{\infty}n\pars{tx^{2}}^{n}}
_{\ds{=\ {tx^{2} \over \pars{1 - tx^{2}}^{2}}}}\ \dd t
\\[5mm] & =
x + x^{3}\
\underbrace{\int_{0}^{1}{\root{1 - t} \over
\pars{1 - tx^{2}}^{2}}\,\dd t}
_{\ds{=\ {-x + \arcsin\pars{x}/\root{1 - x^{2}} \over x^{3}}}} =
\bbx{\arcsin\pars{x} \over \root{1 - x^{2}}}
\end{align}
A: Another answer. We use formulas
$$\int_0^{\pi/2}\sin^{2n+1}sds=\frac{(2n)!!}{(2n+1)!!},$$
$$\sum_{k=0}^{\infty }{\left. {{z}^{2 k+1}}\right.}=\frac{z}{1-{{z}^{2}}},\quad |z|<1.$$
Then
$$f(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}(2n)!!}{(2n+1)!!}\\=\sum_{n=0}^{\infty}x^{2n+1}\int_0^{\pi/2}\sin^{2n+1}sds\\
=\int_0^{\pi/2}\left(\sum_{n=0}^{\infty}(x\sin s)^{2n+1} \right)ds\\=
\int_0^{\pi/2}\frac{x\sin s}{1-x^2\sin^2s}ds\\={\frac {1}{\sqrt {1-x^2}}\arctan \left( {\frac {x}{\sqrt {1-x^2}}} \right) }
$$
We get
$$f\left(\frac{1}{\sqrt2}\right)=\sqrt2\arctan1=\frac{\sqrt2\pi}{4}$$
