Polynomial and product expression of sinx for the 5th power!

Question

Recently I saw a proof of $\frac{\pi^2}6=1+1/2^2+1/3^2+1/4^2+...$ using the taylor expansion of $\sin x$ and the product expansion of $\sin x (x(1-x^2/\pi^2)(1-x^2/(3\pi)^2)...)$ and comparing the third power terms. So I wanted to see what could happen when I did the same for the fifth power and it got messy and I can't find a pattern to make the calculations easier. I know that $\frac{\pi^4}{90}=1+1/2^4+1/3^4+...$ and I couldn't make the fourth power to appear anywhere. For example, I got $\frac14+\frac19+\frac1{4\cdot9}+\frac1{16}+\frac1{4\cdot16} +\frac1{9\cdot16}+...$ and I played around with it a bit and nothing. Does anyone know what the problem appears to be and how or if I can find a nice result? Also, sorry for not writing the equation in the right format, it's been a long time and I forgot how to do it or where to look. If anyone with that as well and I will appreciate it very much. Thank you!

Answer 1

The third degree term is obtained as the sum of the coefficients of $x^2$ in all factors, i.e.

$$-\sum_{k=1}^\infty\frac1{k^2\pi^2}$$ and equate to $-\dfrac1{3!}$.

For the fifth degree term, you can sum all $x^4$ terms obtained from the products of two factors. Hence

$$\sum_{k=1}^\infty\sum_{j=i+1}^\infty\frac1{k^2\pi^2}\frac1{j^2\pi^2}.$$

Using symmetry, this can be written as

$$\frac12\left(\sum_{k=1}^\infty\sum_{j=1}^\infty\frac1{k^2\pi^2}\frac1{j^2\pi^2}-\sum_{k=1}^\infty\frac1{k^4\pi^4}\right)=\frac12\left(\frac1{36}-\sum_{k=1}^\infty\frac1{k^4\pi^4}\right).$$

Equating to $\dfrac1{5!}$, we get

$$\sum_{k=1}^\infty\frac1{k^4\pi^4}=\frac1{36}-\frac2{120}=\frac1{90}.$$