Multiplication of power series explanation Suppose I have two complex valued functions $f(z) = e^z$ and $g(z) = \frac{1}{z^2 + 1}$ How can I derive a series expansion for $f(x)g(x) = \frac{e^z}{z^2 + 1}$ using multiplication of power series? Note that $e^z = \sum^{\infty}_{n=0}{\frac{z^n}{n!}}$ and we can write $\frac{1}{z^2 + 1} = \frac{-1}{1-(z^2 + 2)}$ so that $\frac{-1}{1-(z^2 + 2)} = -\sum^{\infty}_{n=0}{(z^2 + 2)^n}$
 A: The power series of $e^z=\sum_{n\geq 0}b_nz^n$ has infinite radius of convergence.
Now, actually, you should write
$$
\frac{1}{1+z^2}=\frac{1}{1-(-z^2)}=\sum_{n\geq 0}(-z^2)^n=\sum_{n\geq 0}(-1)^nz^{2n}=\sum_{n\geq 0}c_nz^n
$$
which has radius of convergence $1$.
It follows that $\frac{e^z}{1+z^2}$ can be developed as a power series
$$
\frac{e^z}{1+z^2}=\sum_{n\geq 0}a_nz^n
$$
with radius of convergence $\geq 1$ (consequence of the fact that the Cauchy product of two absolutely convergent series converges absolutely, itself a consequence of Mertens' theorem).
The coefficients $a_n$ are computed by convolution
$$
a_n=\sum_{k=0}^{n}b_{n-k}c_k=\sum_{0\leq 2k\leq n}\frac{(-1)^k}{(n-2k)!}.
$$
A: The general way to multiply power series is the generalization of the way we multiply polynomials: if $f(z)=\sum_{n=0}^\infty a_nz^n$ and $g(z)=\sum_{n=0}^\infty b_nz^n$ then:
$$f(z)g(z)=\sum_{n=0}^\infty\left(\sum_{k=0}^na_kb_{n-k}\right)z^n$$
(try this for polynomials and you'll see that it works)
In you case, $e^z = \sum^{\infty}_{n=0}{\frac{z^n}{n!}}$ and $\frac{1}{1+z^2} =\sum^{\infty}_{n=0}(-1)^nz^{2n}$. Hence:
$$\frac{e^z}{z^2+1}=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{1}{k!}b_{n-k}\right)z^n, \hspace{5pt} b_k=\begin{cases}(-1)^m&k=2m\\0&k=2m+1\end{cases}$$
