Obtaining $\zeta(4)$ from specific derived expression

Question

I started from the expansion of cot(x) by the Mittag-Leffler theorem, where:

$$\cot(x) = \frac{1}{x} + \sum_{k = 1}^{\infty} \frac{2x}{x^{2} - k^2 \pi^2}.$$

After splitting up $\cot(x)$, taking the derivative of both sides, and some algebra, I eventually was left with the following expression:

$$-1 = 6\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} + 4x^2 \left [ \sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] ^2 - 4x^2\sum_{k = 1}^{\infty} \frac{1}{({x^{2} - k^2 \pi^2})^2} $$

where $x = 0$ gives $\zeta(2)$. I am now trying to obtain the exact value for $\zeta(4)$ through this expression. To do this, I am attempting to once again take the derivative and then set $x = 0$. My work is shown below.

$$-1 = 6\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} + 4x^2 \left [ \sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] ^2 - 4x^2\sum_{k = 1}^{\infty} \frac{1}{({x^{2} - k^2 \pi^2})^2} \Longrightarrow $$

$\Longrightarrow 0 = -12x\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(x^{2} - k^2 \pi^2)^2} + 8x\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] ^2 + 8x^2\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] \times \frac{d}{dx}\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] - 8x \left [\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(x^{2} - k^2 \pi^2)^2} \right ] \left [ \displaystyle\sum_{k = 1}^{\infty} \frac{-4x}{(x^{2} - k^2 \pi^2)^3} \right ] \Longrightarrow $

Dividing by $x$:

$\Longrightarrow0 =-12\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(x^{2} - k^2 \pi^2)^2} + 8\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] ^2 + \\8x\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] \times \frac{d}{dx}\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{x^{2} - k^2 \pi^2} \right ] - 8x \left [\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(x^{2} - k^2 \pi^2)^2} \right ] \left [ \displaystyle\sum_{k = 1}^{\infty} \frac{-4}{(x^{2} - k^2 \pi^2)^3} \right ] \Longrightarrow $

Setting $x = 0$:

$\Longrightarrow 0 = -12\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(- k^2 \pi^2)^2} + 8\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{- k^2 \pi^2} \right ] ^2 \Longrightarrow $

Multiplying everything by $\pi^4$:

$\Longrightarrow 0 = -12\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(- k^2 )^2} + 8\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{- k^2 } \right ] ^2 \Longrightarrow $

$\Longrightarrow 0 = 3\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(k^2 )^2} - 2\left [ \displaystyle\sum_{k = 1}^{\infty} \frac{1}{ k^2 } \right ] ^2 \Longrightarrow $

Solving for ζ(4):

$\Longrightarrow 0 = 3\displaystyle\sum_{k = 1}^{\infty} \frac{1}{(k^2 )^2} - 2\left [ \frac{\pi^2}{6} \right ] ^2 \Longrightarrow $

$\Longrightarrow \displaystyle\sum_{k = 1}^{\infty} \frac{1}{k^4 } = \frac{\pi^4}{54}$

However, my answer is wrong. The correct result would be $\frac{\pi^4}{90}$. Any help on this would be greatly appreciated.

Answer 1

Expanding the internal passages:

\begin{align} \frac{\text{d}}{\text{d}x} \left(4x^2\left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2\right) & = 8x \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2 + 4x^2 \frac{\text{d}}{\text{d}x} \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2 \\\\ & = 8x \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2 + 8x^2 \sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\frac{\text{d}}{\text{d}x}\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2} \\\\ & = 8x \left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2 -16x^3 \left(\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right) \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2} \end{align}

Also

\begin{align} \frac{\text{d}}{\text{d}x} \left(-4x^2 \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2}\right) & = -8x \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2} - 4x^2 \frac{\text{d}}{\text{d}x}\left(\sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2}\right) \\\\ & = -8x \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2} + 16x^3 \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^3} \end{align}

When you divide by $x$, it remains

$$0 = -12 \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2} + 8\left[\sum_{k = 1}^{+\infty} \frac{1}{x^2 - k^2\pi^2}\right]^2- 8 \sum_{k = 1}^{+\infty} \frac{1}{(x^2 - k^2\pi^2)^2}$$

Setting $x = 0$ then

$$0 = -12 \sum_{k = 1}^{+\infty} \frac{1}{(k^2\pi^2)^2} + 8\left[\sum_{k = 1}^{+\infty} \frac{1}{-k^2\pi^2}\right]^2 -8 \sum_{k = 1}^{+\infty} \frac{1}{(k^2\pi^2)^2} $$

Summing the extreme terms and multiplying by $\pi^4$:

$$0 = -20 \sum_{k = 1}^{+\infty} \frac{1}{(k^2)^2}+ 8\left[\sum_{k = 1}^{+\infty} \frac{1}{-k^2}\right]^2$$

Now

$$\sum_{k = 1}^{+\infty} \frac{1}{k^4} = \frac{8}{20} \frac{\pi^4}{36} = \frac{2\pi^4}{360} = \frac{\pi^4}{90}$$

As wanted.