A calculus problem by Julian Poon

The $n$ -derivative of the Riemann Zeta Function is given by:

$\large \displaystyle \zeta^{(n)} (s) = \frac{d^n}{ds^n} \sum_{k=1}^{\infty} \frac{1}{k^s}$

$\color{#20A900} \boxed{ \large \displaystyle \zeta^{(n)} (s) = \sum_{k=1}^{\infty} \frac{(-1)^n [ \ln(k)]^n}{k^s} }$

So:

$\large \displaystyle S = \sum_{n=0}^{\infty} \frac{\zeta^{(n)}(3)}{n!}$

$\large \displaystyle S = \sum_{n=0}^{\infty}\sum_{k=1}^{\infty} \frac{ (-1)^n [\ln(k)]^n}{n! k^3}$

$\large \displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k^3} \sum_{n=0}^{\infty} \frac{ [-\ln(k)]^n}{n!}$

$\large \displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k^3} e^{-\ln(k)}$

$\large \displaystyle S = \sum_{k=1}^{\infty} \frac{1}{k^4}$

$\color{#20A900} \boxed{ \large \displaystyle S = \zeta(4) = \frac{\pi^4}{90} }$

So:

$\color{#3D99F6} \large \displaystyle A = 4, B = 90, \boxed{ \large \displaystyle A+B=94 }$

Using a Taylor Series expansion centered about the point $s={ s }_{ 0 }$ , we obtain

$\zeta \left( s \right) =\sum _{ n=1 }^{ \infty }{ \frac { { \zeta }^{ \left( n \right) }\left( { s }_{ 0 } \right) }{ n! } } { \left( s-{ s }_{ 0 } \right) }^{ n }$

In order to obtain the form as shown in the question, we can take ${ s }_{ 0 }=3$ . Substituting $s=4$ gives ${ \left( s-{ s }_{ 0 } \right) }^{ n }=1$ for all n, which gives

$\zeta \left( 4 \right) =\sum _{ n=1 }^{ \infty }{ \frac { { \zeta }^{ \left( n \right) }\left( 3 \right) }{ n! } }$

It is known that $\zeta \left( 4 \right) =\frac { { \pi }^{ 4 } }{ 90 }$ , hence $A+B=4+90=\boxed{94}$