# Partial Fractions Expansion of $\tanh(z)/z$

I have seen the following formula in papers (without citations) and in Mathematica's documentation about Tanh[]:

$$\frac{\tanh(z)}{8z}=\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2 \pi^2+4z^2}$$

I have no idea how to prove it and I have also encountered in my research similar sums involving, for instance, $\mathrm{coth}$. It would be nice to have a general method for working with these problems; any suggestions?

• Maybe you can try the Taylor Series?
– JSCB
Jul 27, 2012 at 14:14
• As noted, you should consult a textbook in complex analysis. Usually you will see this as a series for $\tan z$, and then a simple change of variables $iz$ will give you a similar one for $\tanh z$. Jul 28, 2012 at 12:36
• Mittag-Leffler Expansion Aug 5, 2022 at 21:36

There is the infinite product representation

$$\cosh\,z=\prod_{k=1}^\infty \left(1+\frac{4z^2}{\pi^2(2k-1)^2}\right)$$

Taking logarithms gives

$$\log\cosh\,z=\sum_{k=1}^\infty \log\left(1+\frac{4z^2}{\pi^2(2k-1)^2}\right)$$

If we differentiate both sides, we have

$$\tanh\,z=\sum_{k=1}^\infty \frac{\frac{8z}{\pi^2(2k-1)^2}}{1+\frac{4z^2}{\pi^2(2k-1)^2}}$$

which simplifies to

$$\tanh\,z=\sum_{k=1}^\infty \frac{8z}{4z^2+\pi^2(2k-1)^2}$$

Note that the infinite product that we started with is the factorization of $\cosh$ over its (imaginary) zeroes.

Here is a related question.

• @jm this is great; does all such formulas rely on manipulating other known products and sums? Isn't there, for instance, a Fourier transform or generating function approach (or something...)? Again, thank you for the support. Jul 28, 2012 at 1:29
• Well, the more systematic route uses polygamma functions/generalized harmonic numbers, but I thought you'd prefer the route that doesn't require special functions. This is the easiest route I know for the tangent, cotangent, and their hyperbolic counterparts. Jul 28, 2012 at 1:35
• Yeah. Probably. Hehe. It's just that I never know how to deal with sums in any other way than comparing it to other sums. Again, thank you for the answer J.M. Jul 28, 2012 at 1:41
• Our answers to this question seem to be paralleled here :-)
– robjohn
Jul 28, 2012 at 13:23
• @rob, hence the upvote. ;) Jul 28, 2012 at 13:49

In this answer, it is shown that for all $z\in\mathbb{C}\setminus\mathbb{Z}$, $$\pi\cot(\pi z)=\frac1z+\sum_{k=1}^\infty\frac{2z}{z^2-k^2}\tag{1}$$ Applying the identity $\tan(x)=\cot(x)-2\cot(2x)$ to $(1)$ gives \begin{align} \pi\tan(\pi z) &=\sum_{k=1}^\infty\frac{2z}{z^2-k^2}-\sum_{k=1}^\infty\frac{2z}{z^2-\frac{k^2}{4}}\\ &=\sum_{k=1}^\infty\frac{8z}{4z^2-(2k)^2}-\sum_{k=1}^\infty\frac{8z}{4z^2-k^2}\\ &=-\sum_{k=1}^\infty\frac{8z}{4z^2-(2k-1)^2}\\ &=\sum_{k=1}^\infty\frac{8z}{(2k-1)^2-4z^2}\tag{2} \end{align} Applying the identity $\tanh(x)=-i\tan(ix)$ to $(2)$ yields $$\pi\tanh(\pi z)=\sum_{k=1}^\infty\frac{8z}{(2k-1)^2+4z^2}\tag{3}$$ Finally, applying the change variables $z\mapsto z/\pi$ to $(3)$ reveals $$\frac{\tanh(z)}{8z}=\sum_{k=1}^\infty\frac{1}{(2k-1)^2\pi^2+4z^2}\tag{4}$$

• how did you get equation $(2)$ from the series sum? May 4, 2017 at 4:40
• @Danny: I've added a couple of lines of explanation. First, we multiply numerator and denominator by $4$. Then note that we are subtracting the terms $\frac{8z}{4z^2-j^2}$ with both even and odd $j$ from the terms with only even $j$. That leaves the negative of the terms with odd $j$.
– robjohn
May 4, 2017 at 12:57

A general approach to find such expansions is a technique based on expanding a meromorphic function in terms of rational fractions by exploiting its poles. A theorem related to this work is called the Mittag-Leffler theorem in complex analysis. See here (starting from page (96)). You can find examples of such expansions for the functions $\tan(z)\,,\sec(z)\, \cot(z)\,, \csc(z) \,.$

• For a somewhat less obscene reference on Mittag-Leffler theorem, see here.
– Did
Aug 24, 2012 at 11:43
• I reffered the OP where he can find the techniques and examples to his question. It is in the heart of my research. Sep 11, 2012 at 22:35
• I cannot open that link(starting from page (96)) it says you need permission. Jan 26, 2020 at 10:34