# An interesting sum

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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.