Taylor series of $1/(e^x-a)$ about 0?

Question

I know what the Taylor series for $\frac{1}{e^x-1}$ about 0 is, and it's a function of Bernoulli numbers.

However, $\frac{1}{e^x-a}$ seems to be much more convoluted. Does anybody know the general term of this series?

If it exists, finding it on Google is challenging, due to the mathematical notation.

Answer 1

So, maybe the Binomial Theorem is needed. See https://en.wikipedia.org/wiki/Binomial_theorem. Actually, we can obtain that \begin{align} (e^x-a)^{-1} &= -a^{-1}\sum_{k=0}^{\infty} \left(\frac{e^x}{a}\right)^k\\ &=-\frac{1}{a} \sum_{k=0}^{\infty} \frac{1}{a^k} \sum_{n=0}^{\infty} \frac{(kx)^n}{n!}\\ &=-\sum_{n=0}^{\infty} \left(\frac{1}{n!} \sum_{k=0}^{\infty} \frac{k^n}{a^{k+1}} \right)x^n. \end{align}

Well, since $$ \frac{1}{1-y}=\sum_{k=0}^{\infty} y^k,$$ we have that $$ y \left(\frac{1}{1-y}\right)'= \sum_{k=0}^{\infty} k y^k.$$ Let $c_0(y)=\frac{1}{1-y}$, and for $n\geq 1$, let $$c_n(y)=y\cdot c_{n-1}(y)'.$$ Then we have $$-\frac{1}{a}\sum_{k=0}^{\infty} \frac{k^n}{a^k} =-\frac{1}{a}c_n\left(\frac{1}{a}\right).$$

The first five terms are $$ \frac{1}{1-a},-\frac{1}{(a-1)^2},-\frac{a+1}{(a-1)^3},-\frac{a (a+4)+1}{(a-1)^4},-\frac{(a+1) (a (a+10)+1)}{(a-1)^5},-\frac{a (a (a (a+26)+66)+26)+1}{(a-1)^6}.$$

Answer 2

Starting from @Aaron Jia's answer $$\frac{1}{e^x-a}=-\sum_{n=0}^{\infty} \left(\frac{1}{n!} \sum_{k=0}^{\infty} \frac{k^n}{a^{k+1}} \right)x^n$$ let $$c_n=\sum_{k=0}^{\infty} \frac{k^n}{a^{k+1}}=\frac{1}{a}\,\Phi \left(\frac{1}{a},-n,0\right) $$ where appears the Hurwitz-Lerch transcendent function.

So,

$$\frac{1}{e^x-a}=\sum_{n=0}^{\infty} -\frac{\Phi \left(\frac{1}{a},-n,0\right)}{a\, n!} x^n$$ and the coefficients are then $$\left( \begin{array}{cc} 0 & \frac{1}{1-a} \\ 1 & -\frac{1}{(a-1)^2} \\ 2 & \frac{-a-1}{2 (a-1)^3} \\ 3 & \frac{-a^2-4 a-1}{6 (a-1)^4} \\ 4 & \frac{-a^3-11 a^2-11 a-1}{24 (a-1)^5} \\ 5 & \frac{-a^4-26 a^3-66 a^2-26 a-1}{120 (a-1)^6} \end{array} \right)$$ to be compared to the expansion $$\frac{1}{1-a}-\frac{x}{(a-1)^2}+\frac{(-a-1) x^2}{2 (a-1)^3}+\frac{\left(-a^2-4 a-1\right) x^3}{6 (a-1)^4}+\frac{\left(-a^3-11 a^2-11 a-1\right) x^4}{24 (a-1)^5}+\frac{\left(-a^4-26 a^3-66 a^2-26 a-1\right) x^5}{120 (a-1)^6}+O\left(x^6\right)$$