Solve for $x, \ \frac{\pi}{5\sqrt{x + 2}} = \frac 12\sum_{i=0}^\infty\frac{(i!)^2}{x^{2i + 1}(2i + 1)!}$

Question

My friend gave me a question that he found on the internet, and he asked me if I knew where to begin, and I simply said, Sorry pal, I have no idea where to begin. That is why I would like to share this question with the MSE.

Solve for $x$, $$\frac{\pi}{5\sqrt{x + 2}} = \frac 12\sum_{i=0}^\infty\frac{(i!)^2}{x^{2i + 1}(2i + 1)!}$$

Thank you in advance.

P.S. Try not to skip too many steps in the working out, and feel free to provide a hint in the end so I can have a little bit of a challenge :)

Answer 1

Hint: I will only show you how to compute the infinite sum.

Let $t = \dfrac{1}{x}$ and $f(t) = \sum\limits_{n=0}^{\infty}\tfrac{(n!)^2}{(2n+1)!}t^{2n+1}$ and observe that $f'(t) =1+ \sum\limits_{n=1}^{\infty}\dfrac{t^{2n}}{\binom{2n}{n}}$ as well as $$\int_{0}^t\tfrac{f'(u)-1}{u}du = \sum\limits_{n= 1}^{\infty}\dfrac{t^{2n}}{2n\binom{2n}{n}} = \dfrac{t}{\sqrt{4-t^2}}\arcsin\frac{t}{2}$$ when the latter is known as Lehmer's identity and is valid when $|t|<2.$

Answer 2

Notice, that for $|z|<1$, $$\sum_{i=1}^\infty\frac{(2z)^{2i}}{\binom{2i}{i}} =\frac1{1-z^2}\left[z^2+\frac{z\arcsin(z)}{\sqrt{1-z^2}}\right]$$ (for a proof see Does the series $ \sum_{i=1}^{\infty}\frac{2^{i-1}(i-1)!}{\prod_{j=1}^{i}(2j+1)}$ converge and if so, to what?) and by integration we get $$\sum_{i=0}^\infty\frac{(2z)^{2i+1}}{(2i+1)\binom{2i}{i}}= \frac{2\arcsin(z)}{\sqrt{1-z^2}}.$$ Then by letting $2z=1/x$, your equation is $$\frac{\pi}{5\sqrt{x + 2}} = \frac 12\sum_{i=0}^\infty\frac{(1/x)^{2i + 1}}{(2i + 1)\binom{2i}{i}}=\frac{\arcsin(1/(2x))}{\sqrt{1-1/(2x)^2}}.$$ Can you take it from here?

P.S. The unique solution should be the golden ratio $x=\varphi=\frac{\sqrt{5}+1}{2}$ (recall that $\sin(\pi/10)=1/(2\varphi)$).

Answer 3

Using the Beta Function integral: $$ \int_0^1t^k(1-t)^k\,\mathrm{d}t=\frac{k!^2}{(2k+1)!} $$ Therefore, $$ \begin{align} \frac12\sum_{k=0}^\infty\frac{k!^2}{(2k+1)!}x^{-2k-1} &=\frac1{2x}\sum_{k=0}^\infty\int_0^1\frac{t^k(1-t)^k}{x^{2k}}\,\mathrm{d}t\\ &=\frac1{2x}\int_0^1\frac{x^2}{x^2-t(1-t)}\,\mathrm{d}t\\[6pt] &=\frac x2\int_0^1\frac1{\left(t-\frac12\right)^2+x^2-\frac14}\,\mathrm{d}t\\ &=\frac x{2\sqrt{x^2-\frac14}}\int_{-\frac1{2\sqrt{x^2-\frac14}}}^{\frac1{2\sqrt{x^2-\frac14}}}\frac1{t^2+1}\,\mathrm{d}t\\ &=\frac{2x}{\sqrt{4x^2-1}}\tan^{-1}\left(\frac1{\sqrt{4x^2-1}}\right)\\[9pt] &=\frac{2x}{\sqrt{4x^2-1}}\sin^{-1}\left(\frac1{|2x|}\right) \end{align} $$