Where Did Zeta Come From?

For $\left| x \right| >1$ , by geometric sum, we have: $\begin{aligned} \sum_{k=1}^\infty\frac1{x^{2k}} &=\frac1{x^2-1}\\ &=\frac12\left(\frac1{x-1}-\frac1{x+1}\right) \end{aligned}$

On differentiating it 4 times, we get: $\sum_{k=1}^\infty\frac{2k(2k+1)(2k+2)(2k+3)}{x^{2k+4}} =12 \left( \frac1{(x-1)^5}-\frac1{(x+1)^5} \right)$

Now, coming to our summation, we have: $\begin{aligned} \sum_{k=1}^\infty\frac{2k(2k+1)(2k+2)(2k+3)\zeta(2k+4)}{4^{2k+4}} &=\sum_{k=1}^\infty\sum_{j=1}^\infty\frac{2k(2k+1)(2k+2)(2k+3)}{(4j)^{2k+4}}\\ &=\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{2k(2k+1)(2k+2)(2k+3)}{(4j)^{2k+4}}\\ &=12\sum_{j=1}^\infty\left(\frac1{(4j-1)^5}-\frac1{(4j+1)^5}\right)\\ &=12\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)^5}\\ &=-12\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^5}+12 \end{aligned}$

The sum we obtained is Dirichlet-Beta function, which is defined as: $\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}$

We'll use the following recurrence formula to evaluate it: $\beta(2n+1) = -\sum_{k=1}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)$ .

Now we can simply evaluate $\beta(1)=\frac{\pi}{4}$ .

Now, we get $\beta(5)=\frac{5{\pi}^5}{1536}$ .

Now, plugging it into our main summation, we get: $\sum _{ k=1 }^{ \infty }{ \frac { 2k\left( 2k+1 \right) \left( 2k+2 \right) \left( 2k+3 \right) \zeta \left( 2k+4 \right) }{ { 4 }^{ \left( 2k+4 \right) } } } =12-\frac { { 5\pi }^{ 5 } }{ 128 }$

Here, $A=12,B=5,C=5,D=128$ . Hence, $\boxed{A+B+C+D=150}$

I used a variant of the standard identity $\sum_{k=1}^\infty \zeta(2k)z^{2k} \; = \; \tfrac12 (1 - \pi z \cot\pi z)$ to note that $F(z) \; = \; \sum_{k=1}^\infty \zeta(2k+4)z^{2k+3} \; =\; \frac{45 - 15 \pi^2 z^2 - \pi^4 z^4 - 45 \pi z \cot\pi z}{90 z}$ and then calculate the sum as $\frac{1}{4^5}F''''(\tfrac14) \; = \; 12 - \tfrac{5}{128} \pi^5$