Taylor expansion of a function I would like to approximate the function
$f(x)=\frac{2x}{1-e^{-2x}}$
analytically for both small and large $x$.
But when I use the formula for the Taylor expansion, I run into the problem that the function and its derivative are not defined for $x=0$. How can I get around this problem?
For large $x$, my idea was to substitute $y:=1/x$ and then expand the function $g(y)=\frac{2}{y(1-e^{-2x})}$ around $y=0$. However, here I run into the same problem as above.
How do you proceed in such a case?
 A: As $e^{-2x} = 1-2x+2x^2+O(x^3)$, we get
\begin{align}
f(x) &= \frac{2x}{1-e^{-2x}} \\
&= \frac{2x}{1-1+2x-2x^2+O(x^3)} \\
&= \frac{2x}{2x-2x^2+O(x^3)} \\
&= \frac{1}{1-(x+O(x^2))} \\
&= 1+x+O(x^2)
\end{align}
Hence, about $x=0$, $f(x)\approx 1+x$.
For large $x$, as $e^{-2x}<<1$, we get $1-e^{-2x}\approx 1$. Hence, for large $x$, $f(x)\approx 2x$/
A: $$ \frac{2x}{1-e^{-2x}} = 2x\frac{1}{1-e^{-2x}} $$
We can then use $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n} $ to get
$$
2x\sum_{n=0}^\infty e^{-2nx}
$$
We then use the Taylor series of $e^x$ to get
$$
\sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{m}2^{m+1}n^{m}x^{m+1}}{m!}=  2x+\sum_{m=1}^\infty \frac{(-1)^{m}2^{m+1}x^{m+1}}{m!}\sum_{n=0}^\infty n^m 
$$
Where we take the m=0 case out for future avoidance of poles. We can then use $\sum_{n=0}^\infty n^m = \zeta (-m) $ then use the reflection formula for the Riemann-Zeta function.
$$
\zeta (-m) = -2^{-m}\pi^{-1-m}sin(\frac{\pi m}{2})\Gamma (m+1)\zeta (1+m)
$$
To get
$$
2x + 2\sum_{m=1}^\infty \frac{(-1)^{m+1} \zeta (2m+2) x^{2m+2}}{\pi^{2m+2}} = 2x+2\sum_{m=2}^\infty (-1)^m \zeta(2m) \frac{x^{2m}}{\pi^{2m}}
$$
This can also be expressed in terms of Bernoulli numbers
$$
2x-\sum_{m=2}^{\infty} \frac{B_{2m}4^m x^{2m}}{(2n)!}
$$
A: Manipulate the denominator to show that
$$\frac{1}{1-e^{-2x}}=\frac 12 (1+\coth(x))$$
$$f(x)=\frac{2x}{1-e^{-2x}}=x(1+\coth(x))=\sum_{m=0}^\infty \frac{B^+_n }{n!}x^n=\sum_{m=0}^\infty \frac{2^n B_n(1)}{n!}x^n$$
