Finding a sum using $\frac{sin(x)}{x}$ taylor series and in terms of its roots

Question

I saw Euler's proof of how the sum of the squares of the harmonic numbers equals $\frac {\pi^2}{6}$ and wanted to see if it works for the $x^4$ term of the taylor expansion as well. So I found that the coefficient of $x^4$ is $\frac {1}{\pi^4} 1(1/4) + (1 + 1/4)(1/9) + (1 + 1/4 + 1/9)(1/16) + ... = \frac {1}{\pi^4} \frac {n(n+1)(2n+1)}{6((n+1)!)^2}$ which should equal $\frac {1}{120}$ according to the Taylor series for $sin x / x$ but it's not working according to wolfram alpha. What am I doing wrong?

Answer 1

What's the Wolfram alpha's answer? Anyway, if the quick computations I made are correct, the coefficient of $x^4$ in the Taylor expansion of $\sin(x)/x$ should be $$ \frac{1}{\pi^4} \sum_{n \geq 1} \sum_{m \geq 1} \frac{1}{n^2 m^2} = \frac{1}{\pi^4} \left( \sum_{n \geq 1} \frac{1}{n^2} \right) \left( \sum_{m \geq 1} \frac{1}{m^2} \right) = \frac{1}{36} $$ but I should check the computation tomorrow (now I'm in rush sorry!)