Differentiate, differentiate , just differentiate #RMM(1)

Relevant wiki: Riemann Zeta Function

Consider the following product using Maclaurin series of $\tan^{-1}x$ .

$\small {\begin{aligned} \prod_{k=1}^n \tan^{-1} \frac x{2k} & = \left(\frac x2 - \frac {x^3}{3\cdot 2^3} + \frac {x^5}{5\cdot 2^5} - \cdots \right) \left(\frac x4 - \frac {x^3}{3\cdot 4^3} + \frac {x^5}{5\cdot 4^5} - \cdots \right) \left(\frac x6 - \frac {x^3}{3\cdot 6^3} + \frac {x^5}{5\cdot 6^5} - \cdots \right) \cdots \left(\frac x{2n} - \frac {x^3}{3\cdot {2n}^3} + \frac {x^5}{5\cdot {2n}^5} - \cdots \right) \\ & = \prod_{k=1}^n \frac {x}{2^k} - \sum_{j=1}^n \prod_{k\ne j}^n \frac {x}{2^k} \cdot \frac {x^3}{3(2j)^3} + O(x^{n+4}) \\ & = \frac {x^n}{(2n)!!} - \frac {x^n}{(2n)!!} \cdot \frac 13 \left(\frac {x^2}{2^2} + \frac {x^2}{4^2} + \frac {x^2}{6^2} + \cdots + \frac {x^2}{(2n)^2}\right) + O(x^{n+4}) \\ & = \frac {x^n}{(2n)!!} - \frac {x^{n+2}}{12(2n)!!}\sum_{k=1}^n \frac 1{k^2} + O(x^{n+4}) \end{aligned}}$

Therefore,

$\begin{aligned} \Omega & = \lim_{n \to \infty} \lim_{x \to 0} \frac {x^n - (2n)!!\color{#3D99F6}\prod_{k=1}^n \tan^{-1}\frac x{2k}}{x^{n+2}} \\ & = \lim_{n \to \infty} \lim_{x \to 0} \frac {x^n - (2n)!! \color{#3D99F6} \left(\frac {x^n}{(2n)!!} - \frac {x^{n+2}}{12(2n)!!}\sum_{k=1}^n \frac 1{k^2} + O(x^{n+4}) \right)}{x^{n+2}} \\ & = \lim_{n \to \infty} \lim_{x \to 0} \frac {\frac {x^{n+2}}{12}\sum_{k=1}^n \frac 1{k^2} - O(x^{n+4})}{x^{n+2}} & \small \color{#3D99F6} \text{Divide up and down by }x^{n+2} \\ & = \lim_{n \to \infty} \lim_{x \to 0} \left(\frac 1{12}\sum_{k=1}^n \frac 1{k^2} - O(x^2) \right) \\ & = \lim_{n \to \infty} \frac 1{12}\sum_{k=1}^n \frac 1{k^2} & \small \color{#3D99F6} \text{Riemann zeta function } \zeta (n) = \sum_{k=1}^\infty \frac 1{k^n} \\ & = \frac {\color{#3D99F6}\zeta(2)}{12} = \frac {\pi^2}{72} & \small \color{#3D99F6} \text{and } \zeta (2) = \frac {\pi^2}6 \end{aligned}$

Therefore, $a+b = 2 + 72 = \boxed{74}$ .

Here I will present my approaches

$\lim_{n\to \infty}\left(\lim_{x\to 0} \left(\dfrac{x^n-\,(2n)!!\tan^{-1}\frac{x}{2}\cdot \tan^{-1}\frac{x}{4}\cdot \cdots \tan^{-1}\frac{x}{2n}}{x^{n+2}}\right)\right)=\,?$

We have that Taylor series of $\begin{aligned} \tan^{-1} (-1\leq z\leq 1) & =\sum_{k=0}^{\infty}\,(-1)^{k}\dfrac{z^{2k+1}}{2k+1} \end{aligned}$ Setting $z=\frac{x}{2}$ we have then $\begin{aligned}\prod_{m=1}^{n} \tan^{-1} \frac{x}{2m} & = \prod_{m=1}^{n}\left(\sum_{k=0}^{\infty}\,(-1)^{k}\dfrac{x^{2k+1}}{ \,(2m)^{2k+1} \cdot 2k+1}\right)\\ & = x^n\prod_{m=1}^{n}\left(\sum_{k=0}^{\infty}\,(-1)^{k}\dfrac{x^{2k}}{ \,(2m)^{2k+1} \cdot 2k+1}\right) \end{aligned}$ and hence the limit as $\Omega = \dfrac{1}{x^2}\left[1-\,(2n)!!\prod_{m=1}^{n}\left(\sum_{k=0}^{\infty}\,(-1)^{k}\dfrac{x^{2k}}{ \,(2m)^{2k+1} \cdot 2k+1}\right)\right]$ Differentiating w.r.t. x (L-Hopital case) using Leibniz product rule for multiple function and plugging $x=0$ we have then $\begin{aligned} \Omega & = \dfrac{\,(2n)!!}{2}\left[\dfrac{2}{3}\prod_{m=1}^{n}\dfrac{1}{2m} \sum_{m=1}^{n} \left(\dfrac{1}{\,(2m)^2}\right) \right] \\ \Omega = & \dfrac{1}{3} \lim_{n\to \infty} \left(\sum_{m=1}^{n}\dfrac{1}{4m^2}\right) = \dfrac{1}{3}\left(\dfrac{\pi^2}{24}\right)= \dfrac{\pi^2}{72}\end{aligned}$

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Alternative solution Set $\begin{aligned} w & = \dfrac{x^n}{\,(2n)!!}\left(\prod_{k=1}^{n}\dfrac{\tan^{-1}\frac{x}{2k}}{\frac{x}{2k}} \right)\end{aligned}$ Therefore, $x^n-\,(2n)!! w = x^n-x^n\prod_{k=1}^{n}\dfrac{\tan^{-1}\frac{x}{2k}}{\frac{x}{2k}}$ Dividing by $x^{n+2}$ we obtain the following $\dfrac{x^n-(2n)!!w }{x^{n+2}}= \dfrac{1}{x^2}\left(1-\prod_{k=1}^{n}\dfrac{\tan^{-1}\frac{x}{2k}}{\frac{x}{2k}}\right)$ As $x\to 0$ the limit to RHS takes the form of $0/0$ . Since the limit of $\displaystyle \prod_{k=1}^{n}\dfrac{\tan^{-1}\frac{x}{2k}}{\frac{x}{2k}}$ is 1. So let us plug $\begin{aligned} y & = \prod_{k=1}^{n}\dfrac{\tan^{-1}\frac{x}{2k}}{\frac{x}{2k}} \\1- \log y& =1-\sum_{k=1}^{n}\left(\log\left(\sum_{m=0}^{n}(-1)^m\dfrac{2k\cdot x^{2m}}{\,(2k)^{2m+1}\cdot (2m+1)}\right)\right)\end{aligned}$ Differentiating denominator we obtain $2x$ numerator as follows $X = - y\sum_{m=0}^{n}\underbrace{\left(\,(-1)^m\dfrac{x^{2m}}{\,\,(2k)^{2m}\,(2m+1)}\right)^{-1}}_{\text{=1 as x}\to 0} \left((-1)^{m+1}\dfrac{\,2(m+1)x^{2m+1}}{(2(m+1))^2\,(2m+3)}\right)$ and hence our limit becomes $\Omega = \lim_{n\to \infty}\left(\dfrac{2x}{2x}\dfrac{1}{3} \sum_{m=1}^{n}\dfrac{1}{4m^2}\right)= \dfrac{1}{3}\dfrac{\pi^2}{4\cdot 6} = \dfrac{\pi^2}{72}$ Thus required answer is $74$ .