Coefficients of power series involving binomial expansion

Question

I want to expand $\frac{e^z}{z+3}$ around $z=1$.

For it's power series around $z=0$, I computed

$$ \sum_{n=0}^\infty a_n z^n \qquad \mbox{with} \ a_n = \sum_{k=0}^n \frac{(-1)^n 3^{-(n+1)}}{n!(k-n)!} . $$

Using the binomial theorem, I now want to expand this power series around $z=1$. To this end, I use the substitution $u:=z-1$, to simply the notation.

$$ (z-1+1)^n = (u+1)^n \\ = \sum_{k=0}^n \binom nk u^k \ . $$

Now, I'm trying to find the new coefficients $b_n$ such that

$$ \sum_{n=0}^\infty a_n z^n = \sum_{n=0}^\infty \sum_{k=0}^n \binom nk u^k = \sum_{n=0}^\infty b_n u^n . $$

Answer 1

We can work directly. For $u=z-1$, $$f(z)=\frac{e^z}{z+3}=\frac{e^{u+1}}{u+4}=\frac{e}{4}e^u\frac{1}{1+\dfrac{u}{4}}=\frac{e}{4}\left(\sum_{n\ge 0}\frac{1}{n!}u^n\right)\left(\sum_{n\ge 0}\frac{(-1)^n}{4^n}u^n\right)\; (|u|<4),$$ and now, we can use the Cauchy product of series, to obtain $$f(z)=\sum_{n\ge 0}b_nu^n.$$

Answer 2

Expanding on Alone's suggestion,

$$ b_n = \sum_{k=0}^n a_n c_{k-n} = \sum_{k=0}^n \frac{(-1)^n 4^{-n}}{(k-n)!} \ . $$

Hence, the expansion around $z=1$ is given by, $$\frac{e^z}{z+3} = \sum_{n=0}^\infty b_n(z-1)^n \ .$$