A calculus problem by Viki Zeta

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{n^2-n-1} \\ & = \sum_{n=1}^\infty \frac 1{\left(n - \frac {1+\sqrt 5}2\right)\left(n - \frac {1-\sqrt 5}2\right)} \\ & = \frac 1{\sqrt 5} \sum_{n=1}^\infty \left({\color{#3D99F6}\frac 1{n - \frac {1+\sqrt 5}2}} - {\color{#D61F06}\frac 1{n - \frac {1-\sqrt 5}2}} \right) \\ & = \frac 1{\sqrt 5} \sum_{n=1}^\infty \left(\frac 1n - {\color{#D61F06}\frac 1{n + \frac {\sqrt 5-1}2}} - \frac 1n + {\color{#3D99F6}\frac 1{n - \frac {\sqrt 5+1}2}} \right) \\ & = \frac 1{\sqrt 5} \left({\color{#D61F06}\psi \left(\frac {\sqrt 5-1}2+1\right) + \gamma} - {\color{#3D99F6}\psi \left(1-\frac {\sqrt 5+1}2 \right) - \gamma} \right) & \small {\color{#3D99F6}\text{See Notes.}} \\ & = \frac 1{\sqrt 5} \left(\psi \left(\frac {\sqrt 5+1}2\right) - \psi \left(1-\frac {\sqrt 5+1}2 \right) \right) \\ & = \frac 1{\sqrt 5} \left(- \pi \cot \left(\frac {\sqrt 5+1}2 \pi \right) \right) & \small {\color{#3D99F6}\text{See Notes.}} \\ & = \frac {\pi \tan \left(\frac {\sqrt 5}2 \pi \right)}{\sqrt 5} \end{aligned}$

$\implies A+B = 5+2 = \boxed{7}$

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Notes: $\psi(\cdot)$ is the digamma function .

• Theorem: $\displaystyle \psi(s+1)=-\gamma+\sum_{k=1}^\infty \left(\dfrac{1}{k}-\dfrac{1}{k+s}\right)$ , where $\gamma$ is the Euler-Mascheroni constant .
• Theorem: $\psi(1-z) - \psi(z) = \pi \cot \pi z$

Another solution:

Note that $\cos(\pi x)=\prod_{n\ge 1}\left(1-\frac{4x^2}{(2n-1)^2}\right)\\\implies \pi \tan(\pi x)=8x\sum_{n\ge 1}\frac{1}{(2n-1)^2-4x^2}$ by taking $\log$ followed by differentiating. Using $4x^2-1=4\implies x=\frac{\sqrt{5}}{2}$ , we find $\sum_{n\ge 1}\frac{1}{n^2-n-1}=\frac{\pi \tan(\pi \frac{\sqrt{5}}{2})}{\sqrt{5}}$ implying the answer to be $5+2=\boxed{7}$ .