I came across the following problem, let $N![z^N]A(z)$ denote the coefficient of an exponential generating function (EGF) $A(z)$. The EGF is similar to an ordinary generating function (OGF) $A'(z)$ except that instead of the series $A'(z)=\sum_0^Na_Nz^N$ for an OGF, we have $A(z)=a_Nz^N/N!$ for an EGF $A(z)$.
For instance, for EGF $A(z)=e^z$, we have $N![z^N]A(z)=N![z^N]e^z=1$, i.e. the EGF coefficients of $e^z$ are $1$ for all $N \in \mathbb{N}$, i.e. the coefficients of the set $ \{z^0/0!,z^1/1!,z^2/2!,…,z^N/N! \}$ are all $1$ for any $N$ given $e^z$. Similarly, for EGF $A(z)=1/(1-z)$, we have $N!z[^N](1/(1-z))=N!$.
Now, given the following EGF $A(z)$:
$$
A(z)=e^z\int^z_0\frac{1-e^{-t}}{t}dt
$$
We are supposed to get $N![z^N]A(z)=H_N$, where $H_N$ is the $N$th harmonic number, i.e.
$$
N![z^N]e^z\int^z_0\frac{1-e^{-t}}{t}dt = H_N
$$
I could not think of a way to prove the above statement. The problem gave a hint that proving this statement involves forming a differential equation for the EGF $H(z)=\sum_{N \geq 0}H_Nz^N/N!$…
Any help?
Best Answer
I think "forming a differential equation" is an overkill. We can simply do $$A(z)=\int_0^z\frac{e^z-e^{z-t}}{t}\,dt\underset{t=z(1-x)}{\phantom{\big[}=\phantom{\big]}}\int_0^1\frac{e^z-e^{zx}}{1-x}\,dx=\sum_{n=1}^{\infty}\frac{z^n}{n!}\int_0^1\frac{1-x^n}{1-x}\,dx=\sum_{n=1}^{\infty}H_n\frac{z^n}{n!}$$ (the last equality, if unknown to you, follows from $(1-x^n)/(1-x)=1+\ldots+x^{n-1}$).
A side note: if we directly multiply the series for $e^z$ and the integral, we get $$A(z)=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\left(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}z^n}{n\cdot n!}\right)=\sum_{n=1}^{\infty}\frac{z^n}{n!}\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k-1}}{k},$$ i.e. another proof of the "frequent" $G_1=H_1$ case of this question of mine.