Let $f(z)= \sum_{n=0}^{\infty}a_nz^n$ be an entire function. If $\sum_{n=-1}^{\infty}b_nz^n$ is the Laurent series of $\frac{f(z+1)}{z}$ centered on $z=0,$ show that $$b_n = \sum_{k=n+1}^{\infty}\binom{k}{n+1}a_k, n \ge -1.$$
$\frac{f(z+1)}{z} = \sum_{n=-1}^{\infty}b_nz^n \to f(z+1) =\sum_{n=-1}^{\infty}b_nz^{n+1}$
I don't know what to do.
Thanks in advance for any help.
By the binomial theorem, $$(1+z)^k=\sum_{n=0}^k\binom kn z^n\overset{\color{blue}*}=\sum_{n=0}^\infty \binom kn z^n$$.
So we get $$\dfrac {f(z+1)}z=\dfrac {\sum_{k\ge0}a_k(1+z)^k}z=\dfrac {\sum_{k\ge0}a_k(\sum_{n\ge0}\binom kn z^n)}z=\sum_k\sum_n a_k\binom kn z^{n-1}=\sum_n\sum_k a_k\binom kn z^{n-1}=\sum_{n=-1}^\infty \sum_{k\ge0}a_k\binom k{n+1}z^n\overset{\color{blue}*}=\sum_{n=-1}^\infty \sum_{k=n+1}^\infty a_k\binom k{n+1}z^n$$,
$\color {blue}*$ because of the convention that for $k\lt n$, we have $\binom kn=0$.