Power 4

Relevant wiki: Polylogarithm

For $-1< x < 1$ , we have:

$\begin{aligned} \sum_{n=0}^\infty x^n & = \frac 1{1-x} & \small {\color{#3D99F6}\text{Differentiate both sides}} \\ \sum_{n=1}^\infty nx^{n-1} & = \frac 1{(1-x)^2} & \small {\color{#3D99F6}\text{Multiply both sides by }x} \\ \sum_{n=1}^\infty nx^n & = \frac x{(1-x)^2} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^2x^{n-1} & = \frac {1+x}{(1-x)^3} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^2x^n & = \frac {x(1+x)}{(1-x)^3} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^3x^{n-1} & = \frac {1+4x+x^2}{(1-x)^4} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^3x^n & = \frac {x+4x^2+x^3}{(1-x)^4} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^4x^{n-1} & = \frac {1+11x+11x^2+x^3}{(1-x)^5} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^4x^n & = \frac {x+11x^2+11x^3+x^4}{(1-x)^5} & \small {\color{#3D99F6} \text{Put }x = \frac 13} \\ \sum_{n=1}^\infty \frac {n^4}{3^n} & = \frac {\frac 13+ \frac {11}{3^2}+\frac{11}{3^3}+\frac 1{3^4}}{\left(1-\frac 13 \right)^5} \\ & = 15 \end{aligned}$

$\implies a+b = 15+1 = \boxed{16}$ .

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• Alternative solution
• Similar problem

see this Polylogarithm