A digamma series

Let $\psi$ denote the digamma function and $\phi$ the golden ration. Prove that:

$$\sum_{n=1}^{\infty}\frac{\psi(n+\phi) - \psi \left ( n- \frac{1}{\phi} \right )}{n^2+n-1}= \frac{\pi^2}{2\sqrt{5}} + \frac{\pi^2}{\sqrt{5}} \tan^2 \left ( \frac{\pi \sqrt{5}}{2} \right )+ \frac{4\pi }{5}\tan \frac{\pi \sqrt{5}}{2}$$

(I. Mezo, China)

Solution (Ramya Dutta/ India)

Using the digamma identity

$$\begin{equation} \psi (z) = -\gamma + \sum_{n=0}^{\infty} \left ( \frac{1}{n+1} - \frac{1}{n+z} \right ), \;\; z \in \mathbb{C} \setminus \mathbb{Z} \end{equation}$$

we may re-write the nominator as:

\begin{align*}
\psi(n+\phi) - \psi \left ( n - \frac{1}{\phi} \right ) &\overset{1}{=}- \sum_{m=0}^{\infty}\left ( \frac{1}{m+n+\phi} - \frac{1}{m+n - \frac{1}{\phi}} \right ) \\
 &= \left ( \phi + \frac{1}{\phi} \right )\sum_{m=n}^{\infty}\frac{1}{m^2+m-1}\\
 &=\sqrt{5}\sum_{m=n}^{\infty}\frac{1}{m^2+m-1}
\end{align*}

Therefore if $\mathcal{S}$ denotes the initial series then:

\begin{align*}
\mathcal{S} &=\sum_{n=1}^{\infty}\frac{\psi (n+\phi) - \psi (n -\frac{1}{\phi})}{n^2+n-1} \\
 &= \sqrt{5} \sum_{n=1}^{\infty}\sum_{m=n}^{\infty}\frac{1}{\left ( m^2+m-1 \right )\left ( n^2+n-1 \right )}\\
 &= \frac{\sqrt{5}}{2}\left [ \sum_{n=1}^{\infty}\frac{1}{\left ( n^2+n-1 \right )^2} + \left ( \sum_{n=1}^{\infty}\frac{1}{n^2+n-1} \right )^2 \right ]
\end{align*}

For the second sum we have successively:

\begin{align*}
\sum_{n=1}^{\infty}\frac{1}{n^2+n-1} &=\sum_{n=1}^{\infty}\frac{1}{\left ( n+\phi \right )\left ( n - \frac{1}{\phi} \right )} \\
 &=- \frac{1}{\sqrt{5}}\sum_{n=1}^{\infty}\left ( \frac{1}{n+\phi} - \frac{1}{n - \frac{1}{\phi}} \right ) \\
 &= \frac{1}{\sqrt{5}} \sum_{n=1}^{\infty} \left ( \frac{1}{n+\phi} + \frac{1}{-n-1+\phi} \right )\\
 &= -\frac{1}{\sqrt{5}}\left ( - \frac{1}{\phi} - \frac{1}{-1+\phi}+ \sum_{n=-\infty}^{\infty}\frac{1}{n+\phi} \right ) \\
 &= 1- \frac{\pi \cot \pi \phi}{\sqrt{5}}
\end{align*}

since $\sum \limits_{n=-\infty}^{\infty}\frac{1}{z+a}= \pi \cot \pi z , \;\; \mathbb{C}\setminus \mathbb{Z}$. 

For the first sum we have 

\begin{align*}
\sum_{n=1}^{\infty}\frac{1}{\left ( n^2+n-1 \right )^2} &=\frac{1}{5}\sum_{n=1}^{\infty}\left ( \frac{1}{\left ( n+\phi \right )^2} - \frac{2}{\left ( n+\phi \right )\left ( n- \frac{1}{\phi} \right )} + \frac{1}{\left ( n - \frac{1}{\phi} \right )^2} \right ) \\
 &= -\frac{2}{5}\left ( 1 - \frac{\pi \cot \pi \phi}{\sqrt{5}}  \right ) + \frac{1}{5}\sum_{n=1}^{\infty}\frac{1}{\left ( n+\phi \right )^2} + \frac{1}{5}\sum_{n=1}^{\infty}\frac{1}{\left ( n+1-\phi \right )^2}\\
 &= \frac{2\pi \cot \pi \phi}{5\sqrt{5}} - 1 + \frac{\psi^{(1)}(\phi) + \psi^{(1)}(1-\phi)}{5}\\
 &= \frac{2\pi \cot \pi \phi}{5\sqrt{5}} - 1 + \frac{\pi^2}{5\sin^2 \pi \phi}
\end{align*}

because of the trigamma reflection formula.

Hence:

\begin{align*}
\mathcal{S} &=\frac{\sqrt{5}}{2} \left ( - \frac{8\pi \cot \pi \phi}{5\sqrt{5}}+ \frac{\pi^2}{5}\left ( \cot^2 \pi \phi + \csc^2 \pi \phi\right ) \right )\\
 &= -\frac{4\pi \cot \pi \phi}{5}+ \frac{\pi^2}{2\sqrt{5}}+ \frac{\pi^2 \cot^2 \pi \phi}{\sqrt{5}}\\
 &=\frac{\pi^2}{2\sqrt{5}} + \frac{\pi^2}{\sqrt{5}} \tan^2 \frac{\pi \sqrt{5}}{2} + \frac{4\pi}{5} \tan \frac{\pi \sqrt{5}}{2}
\end{align*}