91影视

Replace x with ix in the expression \(\frac{e^x - e^{-x}}{2}\): $$ \frac{e^{ix} - e^{-ix}}{2} $$

Use Euler's formula to express the complex exponentials as a combination of sine and cosine functions: $$ \frac{\cos(x) + i\sin(x) - \cos(-x) - i\sin(-x)}{2}=\frac{2i \sin x}{2} $$ Simplify the expression to get the sine function: $$ i \sin(x) $$

Recall the power series expansion for sine function: $$ \sin(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}x^{2k+1} $$

Recall the infinite product representation for the sine function using the Weierstrass product formula: $$ \sin(x) = x\prod_{k=1}^{\infty}\left(1-\frac{x^2}{\pi^2k^2}\right) $$

From the power series representation, we notice that the coefficient of \(x^2\) is 0 and the coefficient of \(x^4\) is \(\frac{-1}{3!}\). Using the infinite product representation, we can relate these coefficients to the sum of reciprocal of squares and fourth powers. For the coefficient of \(x^2\) from the product representation, we notice that the sum of all terms having \(x^2\) is: $$ \sum_{k=1}^{\infty}\left(-\frac{x^2}{\pi^2k^2}\right) = 0 $$ Multiplying by \(-\pi^2\), we get: $$ \sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6} $$ For the coefficient of \(x^4\) from the product representation, the sum of all terms having \(x^4\) is: $$ \sum_{k=1}^{\infty} \sum_{j=1}^{k-1} \left(-\frac{x^4}{\pi^4k^2j^2}\right) = \frac{-1}{3!} $$ Multiplying by \(-\frac{\pi^4}{4!}\), we have: $$ \sum_{k=1}^{\infty} \sum_{j=1}^{k-1} \frac{1}{k^2j^2} = \frac{\pi^4}{90} $$ which can be rearranged as $$ \sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90} $$