# How to prove this identity on exponential generating function of harmonic numbers

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?

• Your expression for the EGF of the harmonic number can be written in terms of standard functions as $A(z) = e^z (\gamma+\Gamma(0,z)+\log(z))$, where $\gamma$ is Euler's gamma and $\Gamma(0,z)$ is the imcomplete gamma function. Jun 1, 2020 at 19:49

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.

• thanks, just to ask, could you explain how the third equality happened ? Jun 1, 2020 at 3:50
• @LinkL: we substitute $e^z=\sum\limits_{n=0}^{\infty}\frac{z^n}{n!}$ and $e^{zx}$ similarly; the terms with $n=0$ cancel. Jun 1, 2020 at 3:51

Here is a proof involving forming a differential equation.

Let $$H_n$$ be the $$n^\text{th}$$ harmonic number. In particular, $$H_0=0$$ and $$H_n=\frac1n+H_{n-1}$$ for $$n\geq 1$$.

Define the series $$B(t)=\sum_{n\geq 0} H_{n+1}\frac{t^n}{n!}$$. Let $$A(z) = \int_0^z B(t)dt$$. One observation is that: $$A(t)=\sum_{n\geq 0} H_n\frac{t^n}{n!}\label{1} \tag{1}.$$

With that in mind: \begin{align*} B(t) &= \sum_{n\geq 0} H_{n+1}\frac{t^n}{n!} \\ &= \sum_{n\geq 0}\left(\frac1{n+1} + H_n\right)\frac{t^n}{n!} \\ &= t^{-1}\sum_{n\geq 1}\frac{t^n}{n!} + \sum_{n\geq 0}H_n\frac{t^n}{n!} \\ &= t^{-1}(e^t-1) + \int_0^t B(s)ds. \\ \end{align*}

In other words, $$A'(t)-A(t) = t^{-1}(e^t-1)$$. This is a $$1^\text{st}$$ order linear ODE.

By multiplying both sides by the integrating factor $$e^{-t}$$, we get: \begin{align*} [e^{-t}A(t)]' = t^{-1}(1-e^{-t}). \end{align*} Integrate both sides from $$t=0$$ to $$t=z$$ and rearrange to find: $$A(z)=e^z\int_0^z \frac{1-e^{-t}}{t}dt.$$

Using (\ref{1}), we can see that $$n![z^n]A(z)=H_n$$.