Using Taylor series to show that $\frac{h}{2} \cdot \frac{e^h+1}{e^h-1} \approx 1 + \frac{h^2}{12} + \cdots$

Question

How would I use the Taylor series to show the following?

$$\frac{h}{2}\cdot\frac{e^h+1}{e^h-1}\approx1+\frac{h^2}{12}+\cdots$$

Answer 1

$$\dfrac{h}{2}\dfrac{e^h+1}{e^h-1}=\dfrac{h}{2}\dfrac{2+h+\dfrac{h^2}{2!}+\cdots}{h+\dfrac{h^2}{2!}+\cdots}\approx\dfrac{h}{2}\Big[\dfrac{2}{h}+\dfrac{h}{6}+\cdots\Big]\approx 1+\dfrac{h^2}{12}+\cdots$$

Answer 2

$$\frac{h}{2} \left( \frac{e^h+1}{e^h-1} \right) = \frac{h}{2} \, \coth \left( \frac{h}{2} \right) = 1 + \frac{h^2}{12} - \frac{h^4}{720} + \cdots$$

where the Maclaurin expansion for the hyperbolic cotangent was used. Using SymPy instead:

>>> from sympy import *
>>> h = Symbol('h')
>>> f = (h/2) * coth(h/2)

The 4th order expansion is

>>> f.series(h,0,5)
     2     4        
    h     h     / 5\
1 + -- - --- + O\h /
    12   720        

The 12th order expansion is

>>> f.series(h,0,13)
     2     4      6        8        10              12            
    h     h      h        h        h           691*h         / 13\
1 + -- - --- + ----- - ------- + -------- - ------------- + O\h  /
    12   720   30240   1209600   47900160   1307674368000