Closed-form evaluation of$ \sum_{n=1}^{\infty} \frac{1}{n^2 + n - 1}$

Question

While exploring a related mathematical topic, I came across the following infinite series:

$$ \sum_{n=1}^{\infty} \frac{1}{n^2 + n - 1} $$

Numerically, this appears to converge to:

$$ 1 + \frac{\sqrt{5}}{5} \pi \tan\left( \frac{\sqrt{5} \pi}{2} \right) $$

I attempted to prove this identity by factoring the denominator and expressing each term via partial fractions. This leads to terms of the form:

$$ \frac{1}{n^2 + n - 1} = \frac{1}{(n - \alpha)(n - \beta)} $$

where $ \alpha = \frac{-1 + \sqrt{5}}{2}, \quad \beta = \frac{-1 - \sqrt{5}}{2}. $

This allows the series to be rewritten in terms of a telescoping difference involving $(\frac{1}{n - \alpha} - \frac{1}{n - \beta})$, and I suspected that a closed-form might follow from known expansions of digamma or cotangent functions.

I also attempted to use a continued fraction expansion of the cotangent function, but wasn't able to derive the claimed identity.

If anyone knows of a rigorous method to establish this closed form — either using special functions, contour integration, known series identities, or otherwise — I would very much appreciate your insight.

Any approach is welcome; I'm especially interested in how a single identity might admit multiple distinct proofs.

Answer 1

Using partial fraction decomposition, we have $$ \frac{1}{n^2+n-1}=\frac{1}{\sqrt{5}}\left(\frac{1}{n+\frac{1-\sqrt{5}}{2}}-\frac{1}{n+\frac{1+\sqrt{5}}{2}}\right). \tag{1} $$ We can now use the series formula of the digamma function, $$ \psi(z+1)=-\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+z}\right), \tag{2} $$ to rewrite the series $S:=\sum_{n=1}^{\infty} \frac{1}{n^2 + n - 1}$ as \begin{align} S&=\frac{1}{\sqrt{5}}\left(\psi\left(1+\frac{1+\sqrt{5}}{2}\right)-\psi\left(1+\frac{1-\sqrt{5}}{2}\right)\right) \\ &=\frac{1}{\sqrt{5}}\left(\psi\left(\frac{1+\sqrt{5}}{2}\right)+\frac{2}{1+\sqrt{5}}-\psi\left(\frac{1-\sqrt{5}}{2}\right)-\frac{2}{1-\sqrt{5}}\right), \tag{3} \end{align} where we used the recurrence relation $\psi(x+1)=\psi(x)+\frac{1}{x}$ in the last line of $(3)$. Finally, noticing that $\frac{1+\sqrt{5}}{2}=1-\frac{1-\sqrt{5}}{2}$, we can use the reflection formula $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$ to rewrite $(3)$ as \begin{align} S&=\frac{1}{\sqrt{5}}\left(\sqrt{5}+\pi\cot\left(\pi\frac{1-\sqrt{5}}{2}\right)\right) \\ &=1+\frac{\pi}{\sqrt{5}}\tan\left(\frac{\pi\sqrt{5}}{2}\right). \tag{4} \end{align}

Answer 2

One can proceed similarly as in this answer to $\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$ and other sums of quadratic reciprocals.

Start with your partial fraction decomposition: $$ \frac{1}{n^2+n-1} = \frac{1}{\sqrt 5} \left( \frac{1}{n+\frac 12 -\frac 12 \sqrt 5} - \frac{1}{n+\frac 12 + \frac 12 \sqrt 5}\right) $$ Now it is convenient to let the sum start with $n=0$: $$ \begin{align} \sum_{n=0}^\infty \frac{1}{n^2+n-1} &= \frac{1}{\sqrt 5}\sum_{n=0}^\infty \left( \frac{1}{n+\frac 12 -\frac 12 \sqrt 5} - \frac{1}{n+\frac 12 + \frac 12 \sqrt 5}\right) \\ &= \frac{1}{\sqrt 5}\sum_{n=0}^\infty \left( \frac{1}{n+\frac 12 -\frac 12 \sqrt 5} + \frac{1}{-n-\frac 12 - \frac 12 \sqrt 5}\right) \\ &= \frac{1}{\sqrt 5} \lim_{N \to \infty}\sum_{n=-N}^N \frac{1}{n+\frac 12 -\frac 12 \sqrt 5} \\ &= \frac{\pi}{\sqrt 5} \tan\left(\frac{\pi \sqrt 5}{2}\right) \end{align} \, , $$ using the Mittag-Leffler series expansion for the tangent: $$ \pi \tan(\pi z) = \lim_{N \to \infty}\sum_{n=-N}^N \frac{1}{\left(n+\frac 12\right) - z} \, . $$

This confirms your conjecture.

Answer 3

Recall that if $f(z)$ is analytic on $\mathbb{C}$ except at finitely many points $z_1,z_2,\dots,z_k$, none of which is a real integer. Furthermore if there exists $M>0$ such that $|z^2f(z)|\leq M$ for all $|z|>\rho$ for some $\rho>0$. Then, $$\sum_{n\in \mathbb{Z}}f(n)=-\sum_{j=1}^kRes[g(z),z_j]$$ where $\displaystyle{g(z)=\pi\cot(\pi z)f(z)}$ for $z\in \mathbb{C}$.

Now observe that $(n+1)^2-(n+1)-1=n^2+n-1$. It follows that \begin{align*} \sum_{n=-\infty}^{\infty}\frac{1}{n^2+n-1} &=\sum_{n=1}^{\infty}\frac{1}{n^2-n-1}+\sum_{n=0}^{\infty}\frac{1}{n^2+n-1}\\ &=2\sum_{n=0}^{\infty}\frac{1}{n^2+n-1} \end{align*} and therefore \begin{equation} \label{1} \sum_{n=1}^{\infty}\frac{1}{n^2+n-1}=1+\frac{1}{2}\sum_{n=-\infty}^{\infty}\frac{1}{n^2+n-1} \qquad (1) \end{equation} Now, with $\displaystyle{g(z)=\frac{\pi\cot(\pi z)}{z^2+z-1}}$. It's clear that $g(z)$ has simple poles at $\displaystyle{z_1=\frac{-1-\sqrt{5}}{2}}$ and $\displaystyle{z_2=\frac{-1+\sqrt{5}}{2}}$

Thus, $$Res[g(z),z_1]=\frac{\pi\cot(\pi z_1)}{z_1-z_2}=-\frac{\pi}{\sqrt{5}}\tan\left(\frac{\sqrt{5}}{2}\pi\right)$$ and $$Res[g(z),z_2]=\frac{\pi\cot(\pi z_2)}{z_2-z_1}=-\frac{\pi}{\sqrt{5}}\tan\left(\frac{\sqrt{5}}{2}\pi\right)$$ So, we have $$\sum_{n=-\infty}^{\infty}\frac{1}{n^2+n-1}=\frac{2\pi}{\sqrt{5}}\tan\left(\frac{\sqrt{5}}{2}\pi\right)$$ Thus, from $(1)$ we have that $$\sum_{n=1}^{\infty}\frac{1}{n^2+n-1}=1+\frac{\sqrt{5}\pi}{5}\tan\left(\frac{\sqrt{5}}{2}\pi\right)$$