Evaluation of a digamma series involving golden-ratio

Question

Let $\varphi =\frac{1}{2} \left(\sqrt{5}+1\right), a=\tan \left(\frac{\sqrt{5} \pi }{2}\right)$, then how can one prove $$\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+\varphi)-\psi ^{(0)}\left(n-\frac{1}{\varphi}\right)}{n^2+n-1}=\frac{\pi ^2 a^2}{\sqrt{5}}+\frac{4 \pi a}{5}+\frac{\pi ^2}{2 \sqrt{5}}$$ Note that $n^2+n-1=(n+\varphi) \left(n-\frac{1}{\varphi}\right)$. Maybe we should consider the generalized sum i.e. $\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+t)-\psi ^{(0)}(n+s)}{(n+s) (n+t)}$? Any help will be appreciated.

Answer 1

We can use the representation \begin{equation} \psi(a)-\psi(b)=(a-b)\sum_{p=0}^\infty\frac{1}{(p+a)(p+b)} \end{equation} to express the series \begin{align} S(s,t)&=\sum _{n=1}^{\infty } \frac{\psi (n+t)-\psi (n+s)}{(n+s) (n+t)}\\ &=(t-s)\sum _{n=1}^{\infty }\sum_{p=0}^\infty\frac{1}{(p+n+t)(p+n+s)} \frac1{(n+s) (n+t)}\\ &=(t-s)\sum _{n=1}^{\infty }\sum_{k=n}^\infty\frac{1}{(k+t)(k+s)} \frac1{(n+s) (n+t)} \end{align} The case $t=\phi, s=-1/\phi$ is explicitely solved in this question.

For the general case, as shown in the answers, and with $f(n)=1/\left((n+s) (n+t) \right)$, we have \begin{equation} \left[\sum_{n=1}^\infty f(n)\right]^2=2\sum_{n=1}^\infty \sum_{k=n}^{\infty} f(n)f(k) - \sum_{n=1}^{\infty}f(n)^2 \end{equation} then, \begin{equation} S(s,t)=\frac{t-s}{2}\left[\sum _{n=1}^{\infty }\frac1{(n+s) (n+t)}\right]^2+\frac{t-s}{2}\sum _{n=1}^{\infty }\frac1{(n+s)^2 (n+t)^2} \end{equation} With \begin{equation} \frac1{(n+s)^2 (n+t)^2}=\frac{1}{(t-s)^2}\left[\frac{1}{(n+s)^2}+\frac{1}{(n+t)^2}\right]+\frac{2}{(t-s)^3}\left[\frac{1}{n+t}-\frac{1}{n+s}\right] \end{equation} and using again the representations in terms of the polygamma functions given above, we obtain \begin{align} S(s,t)=\frac{1}{2(t-s)}&\left[\psi(t+1)-\psi(s+1)\right]^2\\ &+\frac{1}{2(t-s)}\left[\psi^{(1)}(t+1)+\psi^{(1)}(s+1)\right]\\ &-\frac{1}{(t-s)^2}\left[\psi(t+1)-\psi(s+1)\right] \end{align}

Answer 2

Incomplete answer:

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Since we have

$$\frac1\varphi=\varphi-1$$

It follows that

$$\psi(n+\varphi)-\psi\left(n-\frac1\varphi\right)=\psi(n+\varphi)-\psi(n-\varphi+1)=-\pi\cot(\pi\varphi)+\sum_{k=-n}^{n-1}\frac1{k+\varphi}$$

By recurrence and reflection formulas. We also have

$$\sum_{n=1}^\infty\frac1{n^2+n-1}=1+\frac\pi2\tan\left(\frac{\sqrt5}2\pi\right)$$

so it remains to tackle

$$\sum_{n=1}^\infty\frac1{n^2+n-1}\sum_{k=-n}^{n-1}\frac1{k+\varphi}$$