A calculus problem by Shivam Jadhav

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

$\begin{aligned} \sum_{n=0}^\infty x^n & = \frac 1{1-x} & \small {\color{#3D99F6}\text{Differentiate both sides}} \\ \sum_{n=1}^\infty nx^{n-1} & = \frac 1{(1-x)^2} & \small {\color{#3D99F6}\text{Multiply both sides by }x} \\ \sum_{n=1}^\infty nx^n & = \frac x{(1-x)^2} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^2x^{n-1} & = \frac {1+x}{(1-x)^3} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^2x^n & = \frac {x(1+x)}{(1-x)^3} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^3x^{n-1} & = \frac {1+4x+x^2}{(1-x)^4} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^3x^n & = \frac {x+4x^2+x^3}{(1-x)^4} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^4x^{n-1} & = \frac {1+11x+11x^2+x^3}{(1-x)^5} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^4x^n & = \frac {x+11x^2+11x^3+x^4}{(1-x)^5} & \small {\color{#3D99F6}\text{Differentiate again}} \\ \sum_{n=1}^\infty n^5x^{n-1} & = \frac {1+26x+66x^2+26x^3+x^4}{(1-x)^6} & \small {\color{#3D99F6} \times x \text{ again}} \\ \sum_{n=1}^\infty n^5x^n & = \frac {x+26x^2+66x^3+26x^4+x^5}{(1-x)^6} & \small {\color{#3D99F6} \text{Put }x = \frac 15} \\ \sum_{n=1}^\infty \frac {n^5}{5^n} & = \frac {\frac 15 +\frac {26}{5^2}+\frac {66}{5^3}+\frac {26}{5^4}+\frac 1{5^5}}{\left(1-\frac 15\right)^6} \\ & = \frac {3535}{512} \\ & = \frac {35\cdot 101}{8^3} \end{aligned}$

$\implies M-W = 101-8 = \boxed{93}$ .

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Alternative solution

The following can be done by using polylogaritms,

$Polylog [-5,5] = \frac{3535}{512}$

Hence,

$101-8 = 93$