Complex power series expansion of $\frac{e^z}{1+z}$

I'm trying to find complex power series expansion of $\frac{e^z}{1+z}$ centered at $z=0$ and its radius of convergence. Here is my attempt:

Since $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$, we can divide both terms by $1+z$ to get $e^z = \sum_{n=0}^\infty \frac{z^n}{(1+z)n!}$. Then I can get radius of convergence using the usual Cauchy-Hadamard formula.

Is this correct? Thanks for any help!

• What you obatin is not a power series expansion since there's a $z$ in the denominators. Apr 17 '16 at 16:21
• See math.stackexchange.com/questions/521894/… for a correct approach. Apr 17 '16 at 16:32

In order to get the power series expansion we could multiply the series $e^z$ with $\frac{1}{1+z}$ using the Cauchy product

We obtain \begin{align*} \frac{e^z}{1+z}&=\left(\sum_{k=0}^{\infty}\frac{z^k}{k!}\right)\left(\sum_{l=0}^\infty (-z)^l\right)\tag{1}\\ &=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{(-1)^l}{k!}\right)z^n\tag{2}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\frac{(-1)^{n-k}}{k!}\right)z^n \end{align*}

Comment:

• In (1) we use the series expansion for the exponential function and the geometric series expansion

• In (2) we multiply the series using the Cauchy product formula

We observe the function \begin{align*} \frac{1}{1+z} \end{align*} has a simple pole at $z=-1$. We also know that the exponential function is an entire function, i.e. analytic in $\mathbb{C}$. Since the radius of convergence is the distance from the center $z=0$ to the nearest singularity, we conclude the radius is $1$.

As said, it is false, you need to obtain something like $$\sum_{n=0}^{+\infty} a_n z^n.$$ To answer the question, remark that for $|z|<1$ we have $$\frac{1}{1+z} = \sum_{n=0}^{+\infty} (-1)^n z^n$$ and of course $$e^z = \sum_{n=0}^{+\infty} \frac{z^n}{n!}.$$

Gather the like powers $$\left(1+\frac{1}{1!}z+\frac{1}{2!}z^2+\frac{1}{3!}z^3+\cdots\right)(1-z+z^2-z^3+\cdots) \\ =1+(-1+\frac{1}{1!})z+(1-\frac{1}{1!}+\frac{1}{2!})z^2+(-1+\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!})z^3+\cdots$$ The resulting series is guaranteed to converge in $|z| < 1$ because the first two series converge absolutely in that disk. It can't converge in a disk of radius $r > 1$ because that disk includes the singularity of $\frac{1}{1+z}$ at $z=-1$.