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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?

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    $\begingroup$ 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. $\endgroup$ Commented Jun 1, 2020 at 19:49

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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.

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  • $\begingroup$ thanks, just to ask, could you explain how the third equality happened ? $\endgroup$
    – Link L
    Commented Jun 1, 2020 at 3:50
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    $\begingroup$ @LinkL: we substitute $e^z=\sum\limits_{n=0}^{\infty}\frac{z^n}{n!}$ and $e^{zx}$ similarly; the terms with $n=0$ cancel. $\endgroup$
    – metamorphy
    Commented Jun 1, 2020 at 3:51
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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 $n![z^n]A(z)=H_n$.

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