My first approach was to try and convert the summation into a derived result of the Basel problem. However I couldn't find a link and used a mix of polynomial and trigonometric functions to represent the LHS.
Assumed: $\arctan^2\left(\dfrac{x}{1.2}+\dfrac{x^2}{2.3}+\dfrac{x^3}{3.4}+\cdots\right)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots$
$a_2=\frac{1}{2.2!}$;$a_3=\frac{3}{3.3!}$;$a_4=\frac{5}{3.4!}$;$a_5=\frac{8}{3.5!}$
I am not sure if the derived result is correct. I would appreciate any alternate derivations that either prove or disprove the result. Additionally, I would like to know if the RHS might somehow be connected to the Basel problem.
Assumed
$\arctan^2\left(\dfrac{x}{1.2}+\dfrac{x^2}{2.3}+\dfrac{x^3}{3.4}+\cdots\right)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots$
The resulting coefficients are $a_0=0; a_1=0$(Since LHS is also missing x coefficients) $a_2=\frac{1}{4}; a_3=\frac{1}{6}; a_4=\frac{5}{72}; a_5=\frac{8}{360}$
When $x=1$; the LHS becomes $\arctan^2(1)=\left(\dfrac{\pi}{4}\right)^2=\dfrac{\pi^2}{16}$
Since $a_2=\frac{1}{2.2!}$;$a_3=\frac{3}{3.3!}$;$a_4=\frac{5}{3.4!}$;$a_5=\frac{8}{3.5!}$
Following the pattern, $a_6=\frac{8.11}{25.7.6!}$; $a_7=\frac{88.14}{25.7.9.7!}$ ;$a_8=\frac{8.11.14.17}{25.7.9.11.8!}...$
The RHS becomes:
$\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{5}{72}+\dfrac{8}{360}+\cdots$
Therefore;
$\dfrac{\pi^2}{16}=\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{5}{72}+\dfrac{8}{360}+\dfrac{44}{525}+\dfrac{11}{70875}+\cdots$