# Taylor series of $\frac{1}{1-\exp(z-1)}$

As part ofa voluntary homework assignment, I want to find the Taylor series around 0 for the complex valued function $$f(z):=\frac{1}{1-\exp(z-1)}$$

My approach so far is to express $\exp(z-1)$ and $\frac{1}{1-z}$ as power series. Since $\exp(z)= \sum \frac{z^k}{k!}$ and $\frac{1}{1-z} = \sum{z^k}$, this leads me to $$\frac{1}{1-\exp(z-1)} = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^m}{m!} \right)^k = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^{mk}}{(m!)^k} \right)$$ To obtain a power series around 0, I use the binomial series: $$\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \left( \frac{(z-1)^{mk}}{(m!)^k} \right) = \sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \sum_{n=0}^{mk} \frac{1}{(m!)^k} \binom{mk}{n} z^n \cdot (-1)^{mk-n}$$ However, I am not sure how to proceed at this point. Can the last expression be simplified?

Of course, one could find the solution by using derivatives, but I should find the coefficients of the Taylor series based on the geometric and exponential power series. Am I overlooking something?

• Did you try by separating out $e^{-1}$ from $e^{z-1}$? treating $e^{-1}$ as a fixed constant may help write the power series in powers of $z$. Also try to write few terms explicitly. Sometimes too many summations may be more confusing. Commented Jun 3, 2018 at 16:32

HINT: An easier place to start would be $$\frac{1}{1-e^{z-1}} = \sum_{n=0}^{\infty}\frac{e^{nz}}{e^n}$$ Then expand the exponential $e^{nz}$ in the variable $nz$, instead of raising a power series to an exponent.

$$f(z):=\frac{1}{1-e^{-1}\exp(z)} = \sum_{k=0}^{\infty} e^{-k}e^{zk} = \sum_{k=0}^{\infty} e^{-k} \sum_{m=0}^{\infty} \frac{(zk)^{m}}{m!}$$ Now exchange the summation orders: $$f(z) =\sum_{m=0}^{\infty} \frac{z^m}{m!}\; \sum_{k=0}^{\infty} e^{-k} k^m =\sum_{m=0}^{\infty} \frac{z^m}{m!}\; c_m$$ So the difficulty arises in evaluating the coefficients $c_m = \sum_{k=0}^{\infty} e^{-k} k^m$. Write $$\sum_{k=0}^{\infty} e^{-k} k^m = \lim_{a\to -1}\sum_{k=0}^{\infty} e^{ak} k^m \\ = \lim_{a\to -1}\sum_{k=0}^{\infty} \frac{\partial^m}{\partial a^m}e^{ ak} = \\ = \lim_{a\to -1} \frac{\partial^m}{\partial a^m}\sum_{k=0}^{\infty} e^{ak} = \\ = \lim_{a\to -1} \frac{\partial^m}{\partial a^m}\frac{1}{1-e^{a}}$$ and proceed from there. A few coefficients are (note they do not include the $1/m!$): \begin{align} c_0 &= \frac{e}{e-1} &\\ c_1 &= \frac{e}{(e-1)^2} &\\ c_2 &= \frac{e(1+e)}{(e-1)^3} &\\ c_3 &= \frac{e(1 + 4e + e^2)}{(e-1)^4} &\\ c_4 &= \frac{e (1 + 11 e + 11 e^2 + e^3)}{(e-1)^5} & \end{align} They can be written as a finite sum $c_m = e\sum_{j=0}^{m}{m\brace j}\frac{j!}{\left(e-1\right)^{j+1}}$, see Markus Scheuer's answer.

• Are those $\frac{\delta}{\delta a}$ expressions meant to be functional derivatives or normal partial derivatives? Commented Jun 3, 2018 at 17:51
• @probably_someone Partial derivatives. I had used a wrong "partial" - fixed now. Commented Jun 3, 2018 at 17:54

I think it is more convenient to start with \begin{align*} f(z)=\frac{1}{1-e^{z-1}}&=\sum_{n=0}^\infty e^{(z-1)n} \end{align*} and calculate the Taylor series via \begin{align*} f(z)=\sum_{n=0}^\infty f^{(n)}(0)\frac{z^n}{n!} \end{align*}

We obtain \begin{align*} \color{blue}{\left.\left(\frac{d^k}{dz^k}f(z)\right)\right|_{z=0}}&=\left.\sum_{n=0}^\infty n^ke^{(z-1)n}\right|_{z=0}\tag{1}\\ &=\sum_{n=0}^\infty n^k\left(\frac{1}{e}\right)^n\tag{2}\\ &=\sum_{n=0}^\infty \sum_{j=0}^k{k\brace j}n^{\underline{j}}\left(\frac{1}{e}\right)^n\tag{3}\\ &=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\sum_{n=0}^\infty n^{\underline{j}}\left(\frac{1}{e}\right)^{n-j}\tag{4}\\ &=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\left.\left(\frac{d^j}{dz^j}\frac{1}{1-z}\right)\right|_{z=\frac{1}{e}}\tag{5}\\ &=\sum_{j=0}^k{k\brace j}\left(\frac{1}{e}\right)^j\frac{j!}{\left(1-\frac{1}{e}\right)^{j+1}}\tag{6}\\ &\,\,\color{blue}{=e\sum_{j=0}^{k}{k\brace j}\frac{j!}{\left(e-1\right)^{j+1}}}\\ \end{align*}

Comment:

• In (1) we differentiate $k$ times.

• In (2) we evaluate the expression at $z=0$.

• In (3) we represent $n^k$ with the help of the Stirling numbers of the second kind as sum of falling factorials $n^{\underline{j}}=n(n-1)\cdots(n-j+1)$.

• In (4) we exchange the sums.

• In (5) we use the geometric series expansion.

• In (6) we differentiate $j$ times and evaluate at $z=\frac{1}{e}$.

We conclude \begin{align*} \color{blue}{\frac{1}{1-e^{z-1}}=e\sum_{n=0}^\infty \left(\sum_{j=0}^{n}{n\brace j}\frac{j!}{\left(e-1\right)^{j+1}}\right)\frac{z^n}{n!}} \end{align*}