An interesting sum

I am not an analyst, so I find the sums of infinite series quite mysterious. For example, here are three. The first one is the value of \(\zeta(2)\), very well known, sometimes called the "Basel Problem" and first determined by (of course) Euler: \[ \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}. \] Second, subtracting one from the denominator: \[ \sum_{n=2}^\infty\frac{1}{n^2-1}=\frac{3}{4} \] This sum is easily demonstrated by partial fractions: \[ \frac{1}{n^2-1}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right) \] and so the series can be expanded as: \[ \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}\cdots\right) \] This is a telescoping series in which every term in the brackets is cancelled except for \(1+1/ 2\), which produces the sum immediately.

Finally, add one to the denominator: \[ \sum_{n=2}^\infty\frac{1}{n^2+1}=\frac{1}{2}(\pi\coth(\pi)-1). \] And this sum is obtained from one of the series representations for \(\coth(z)\):

\[ \coth(z)=\frac{1}{z}+2z\sum_{n=1}^\infty\frac{1}{\pi^2n^2+z^2} \]

(for all \(z\) except for when \(\pi^2n^2+z^2=0\)).

I was looking around for infinite series to give my numerical methods students to test their powers of approximation, and I came across [this beauty](https://www.wolframalpha.com/input/?i=sum+1%2F(n^2%2Bn-1)%2C+n%3D1+to+infinity): \[ \sum_{n=2}^\infty\frac{1}{n^2+n-1}=1+\frac{\pi}{\sqrt{5}}\tan\left(\frac{\sqrt{5}\pi}{2}\right). \] This led me on a mathematical treasure hunt through books and all over the internet, until I had worked it out.

My starting place, after googling "sum quadratic reciprocal" was a very nice and detailed post on stackexchange. This post then referred to a previous one started with the infinite product expression for \(\sin(x)\) and turned it (by taking logarithms and differentiating) into a series for \(\cot(x)\).

However, I want an expression for \(\tan(x)\), which means starting with the infinite product form for \(\sec(x)\), which is:

\[ \sec(x)=\prod_{n=1}^\infty\frac{\pi^2(2n-1)^2}{\pi^2(2n-1)^2-4x^2}. \] Making a substitution simplifies the expression in the product: \[ \sec\left(\frac{\pi x}{2}\right)=\prod_{n=1}^\infty\frac{(2n-1)^2}{(2n-1)^2-x^2}. \] Now take logs of both sides:

\[ \log\left(\sec\left(\frac{\pi x}{2}\right)\right)= \sum_{n=1}^\infty\log\left(\frac{(2n-1)^2}{(2n-1)^2-x^2}\right) \]

and differentiate: \[ \frac{\pi}{2}\tan\left(\frac{\pi x}{2}\right)= \sum_{n=1}^\infty\frac{2x}{(2n-1)^2-x^2}. \] Now we have to somehow equate this new sum on the right with our original sum. So let's go back to it.

First of all, a bit of completing the square produces \[ \frac{1}{n^2+n-1}=\frac{1}{\left(n+\frac{1}{2}\right)^2-\frac{5}{4}}=\frac{4}{(2n+1)^2-5}. \] This means that \[ \sum_{n=1}^\infty\frac{1}{n^2+n-1}=\sum_{n=2}^\infty\frac{4}{(2n-1)^2-5}= \frac{2}{\sqrt{5}}\sum_{n=2}^\infty\frac{2\sqrt{5}}{(2n-1)^2-5}. \] We have changed the index from \(n=1\) to \(n=2\) which allows the rewriting of \(2n+1\) as \(2n-1\). This means we are missing a first term. Comparing the final sum with that for \(\tan(x)\) above, we have \[ \sum_{n=1}^\infty\frac{1}{n^2+n-1}=\frac{2}{\sqrt{5}}\left(\frac{\pi}{2}\tan\left(\frac{\pi \sqrt{5}}{2}\right)-\frac{-\sqrt{5}}{2}\right) \] where the last term is the missing first term: the summand for \(n=1\). Simplifying the right hand side produces \[ \sum_{n=1}^\infty\frac{1}{n^2+n-1}=1+\frac{\pi}{\sqrt{5}}\tan\left(\frac{\sqrt{5}\pi}{2}\right). \] Note that the above series for \(\tan(x)\) can be obtained directly, using a general technique discussed (for example) in that fine old text: "A Course in Modern Analysis", by E. T. Whittaker and G. N. Watson. If \(f(x)\) has only simple poles \(a_n\) with residues \(b_n\), then \[ f(x) = f(0)+\sum_{n=1}^\infty\left(\frac{1}{x-a_n}+\frac{1}{a_n}\right). \] Expressing a function as a series of such reciprocals is known as Mittag-Leffler's theorem and in fact the series for \(\tan(x)\) is given there as one of the examples.