Taylor series of $\frac{\cosh(t)-1}{\sinh(t)}$

Question

Please help me to prove that $$\frac{\cosh(t)-1}{\sinh(t)}=\frac{t}{2}\Big(1-\frac{t^2}{12}+\mathcal{O}(t^4)\Big).$$

This is what I wrote: $$\cosh(t)-1=\frac{t^2}{2}+\frac{t^4}{24}+\mathcal{O}(t^6)\quad\text{and}\quad \sinh(t)=t+\frac{t^3}{6}+\mathcal{O}(t^5).$$ But I don't know how to find the result. Thanks

Answer 1

Since your function is an odd function, its Taylor series at $0$ is of the type $a_1t+a_3t^3+a_5t^5+\cdots$ So, all that remains to be proved is that $a_1=\frac12$ and that $a_3=-\frac1{24}$. But\begin{align}\frac{t^2}2+\frac{t^4}{24}+\cdots&=\left(a_1t+a_3t^3+\cdots\right)\left(t+\frac{t^3}6+\cdots\right)\\&=a_1t^2+\left(\frac{a_1}6+a_3\right)t^4+\cdots,\end{align}you get that, indeed, $a_1=\frac12$ and $a_3=-\frac1{24}$.

Answer 2

$\frac{\cosh(t)-1}{\sinh(t)}=\tanh\frac{t}{2}$ and since $\tan t=-\frac{d}{dt}\log\cos t$ and $$ \cos(t)=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\tag{Weierstrass product}$$ we have: $$ \tan(t) = \sum_{n\geq 1}\frac{2(4^n-1)\zeta(2n)}{\pi^{2n}}t^{2n-1} \tag{A}$$ $$ \tanh(t) = \sum_{n\geq 1}\frac{2(4^n-1)\zeta(2n)(-1)^{n+1}}{\pi^{2n}}t^{2n-1} \tag{B}$$ $$ \tanh\frac{t}{2} = \sum_{n\geq 1}\frac{4(4^n-1)\zeta(2n)(-1)^{n+1}}{(2\pi)^{2n}}t^{2n-1}=\color{red}{\frac{t}{2}-\frac{t^3}{24}+O(t^5)} \tag{C}$$ since $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$. This is usually exploited in the opposite direction: since the Taylor series at the origin of $\tan t$ is what it is, $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$.