# 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?

• Idea: differentiate, divide by $x$, integrate, keep fiddling to retrieve original function in a differential equation. No pencil and paper handy so I can't tell if this strategy works ... – Ethan Bolker Jan 28 at 17:34

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 $$f\left(\frac{1}{\sqrt2}\right)=\sqrt2\arctan1=\frac{\sqrt2\pi}{4}$$

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

• How do we predict that the given series is the expansion of $\frac{arcsin(x)}{\sqrt{1-x^2}}$ ? – Akshat Joshi Jan 29 at 2:24
• In my opinion, this answer only works if you already know the answer. How would you handle this problem if you didn't happen to know the series expansion of $\arcsin(x)/\sqrt{1 - x^2}$? – John Barber Jan 29 at 4:35


$$\ds{\left.\vphantom{\Large A}\mrm{f}\pars{x}\right\vert_{\ x\ \in\ \left[0,1\right)} \equiv x + {2x^{3} \over 1 \times 3} + {2 \times 4x^{5} \over 1 \times 3 \times 5} + {2 \times 4 \times 6x^{7} \over 1 \times 3 \times 5 \times 7} + \cdots:\ {\LARGE ?}}$$.

$$\underline{\large\texttt{An Explicit Evaluation:}}$$ \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}

• Very nice use of Beta integral +1 – Paramanand Singh Jan 29 at 8:01
• Thanks for the help. I am still in high school and not introduced to the gamma function. Any alternate way of doing it? – Akshat Joshi Jan 29 at 12:16

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,

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