partial fractionssequences-and-seriestrigonometry
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?
You are certainly aware that the hyperbolic tangent has an asymptote; since no polynomial ever had a horizontal asymptote, it stands to reason that a polynomial will always poorly approximate the qualitative behavior of the hyperbolic tangent.
One viable approach would be to consider the rational functions that come from truncations of the continued fraction for the hyperbolic tangent:
$$\tanh\,z=\cfrac{z}{1+\cfrac{z^2}{3+\cfrac{z^2}{5+\cdots}}}$$
Here for instance are plots comparing $\tanh\,z$ and the convergent
$$R(z)=\cfrac{z}{1+\cfrac{z^2}{3+\cfrac{z^2}{5+\cfrac{z^2}{7+\cfrac{z^2}{9+\cfrac{z^2}{11}}}}}}$$
The left plot shows that the rational approximant and the actual function are almost visually indistinguishable, while the right plot depicts the function $\tanh\,z-R(z)$.
One other possibility you can use in conjunction with rational function approximation is the use of argument reduction; in particular, the identity
$$\tanh\,z=\frac{2\tanh\frac{z}{2}}{1+\tanh^2\frac{z}{2}}$$
is very useful here. To sketch a possible algorithm: scale $z$ by $2^{-k}$, where $k$ is appropriately chosen such that $\dfrac{z}{2^k}$ is "small enough", evaluate the truncated continued fraction at this reduced argument, and then repeatedly apply the double argument identity $k$ times.
You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator: $$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$ We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):
\begin{alignat}{6} & & \color{red}{\dfrac13 -\dfrac49x}&\color{red}{{}+\dfrac{11}{9}x^2} & \color{red}{{}-\dfrac{182}{81}x^3}&+\dotsm\\% 9 + 12x-11x^2+ 2x^3 & \enspace\biggl(\enspace & 3+2x^2 \\[-18mu] & & -3 - 4x& +\dfrac{11}{3}x^2 &{} -\dfrac23 x^3 \\ & & -4x&+ \dfrac{17}3x^2&{} -\dfrac{2}{3}x^3 \\ &&4x&+\dfrac{16}3x^2&{} -\dfrac{44}{9}x^3&+\dotsm\\ &&&\hskip30mu 11x^2&{}-\dfrac{50}{9}x^3&+\dotsm\\ &&&\hskip12mu - 11x^2&{}-\dfrac{132}{9}x^3&+\dotsm\\ &&&&-\dfrac{182}{9}x^3&+\dotsm \end{alignat}
Best Answer
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.