Who's up to the challenge? 30

Calculus Level 5

$\large \displaystyle\sum _{ n=2 }^{ \infty }{ \dfrac { 1 }{ { n }^{ 4 }-{ n }^{ 3 } } } =\dfrac { { \psi }^{ (H) }(U) }{ M } +{ M }_{ 1 }-\frac { { \pi }^{ { U }_{ 1 } } }{ S }$

The equation above holds true for positive integers
$H,U,M,M_1,U_1$ and $S$ . Find $H+U+M+M_1+U_1+S$ .

Notation : $\psi^{(n)}(\cdot)$ denotes the $n^\text{th}$ derivative of the Digamma function .

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The answer is 16.

1 solution

Mark Hennings
Apr 6, 2016

$\begin{array}{rcl} \displaystyle \sum_{n=2}^\infty \frac{1}{n^4 - n^3} & = & \displaystyle \sum_{n=2}^\infty \Big[\frac{1}{n-1} - \frac{1}{n} - \frac{1}{n^2} - \frac{1}{n^3}\Big] \; = \; 1 - \sum_{n=2}^\infty n^{-2} - \sum_{n=2}^\infty n^{-3} \\ & = & 3 - \zeta(2) - \zeta(3) \; = \; \tfrac12 \psi''(1) + 3 - \tfrac16\pi^2 \end{array}$ making the answer $2 + 1 + 2 + 3 + 2 + 6 = \boxed{16}$ .

Who's up to the challenge? 30

The answer is 16.

1 solution

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