Asymptotic expansion of $\sum_{n=1}^\infty\frac{H_n}{n!}z^n$ for $z\to\infty$

Question

What is the asymptotic expansion of $\sum_{n=1}^\infty\frac{H_n}{n!}z^n$ for $z\to\infty$ where $H_n=\sum_{k=1}^n \frac{1}{k}$?

I thought of using the Euler-Mascheroni constant, the fact that $\gamma=\lim_{n\to\infty}(H_n-\log(n))$ and expressing it in form of $Ei(z)$ of which I know the asymptotic expansion of. But I'm not getting anywhere. Can someone help me out and suggest a direction in which way to look? Is there maybe a way to represent the series as an integral?

Answer 1

Using this question and $\gamma=\int_0^1\frac{1-e^{-t}}{t}\,dt-\int_1^\infty\frac{e^{-t}}{t}\,dt$, one gets $$\sum_{n=1}^\infty\frac{H_n}{n!}z^n=e^z\int_0^z\frac{1-e^{-t}}{t}\,dt=e^z\left(\log z+\gamma+\int_z^\infty\frac{e^{-t}}{t}\,dt\right).$$ The last integral has a well-known asymptotics: $$e^z\int_z^\infty\frac{e^{-t}}{t}\,dt\underset{t=z(1+x)}{\phantom{\big[}=\phantom{\big]}}\int_0^\infty\frac{e^{-zx}}{1+x}\,dx\asymp\sum_{n=0}^{(\infty)}\frac{(-1)^n n!}{z^{n+1}}\qquad(z\to+\infty)$$ (obtained using Watson's lemma, or simply integration by parts).

Answer 2

Let $$ f(z)=\sum_{n=0}^\infty\frac{H_n}{n!}z^n\tag1 $$ Then $$ \begin{align} f'(z) &=\sum_{n=0}^\infty\frac{H_{n+1}}{n!}z^n\tag{2a}\\ &=\sum_{n=0}^\infty\frac{H_n}{n!}z^n+\sum_{n=0}^\infty\frac{z^n}{(n+1)!}\tag{2b}\\ &=f(z)+\frac{e^z-1}z\tag{2c} \end{align} $$ Explanation:
$\text{(2a)}$: take the derivative term by term, then substitute $n\mapsto n+1$
$\text{(2b)}$: $H_{n+1}=H_n+\frac1{n+1}$
$\text{(2c)}$: apply $(1)$ to the left hand sum and
$\phantom{\text{(2c):}}$ the series for $e^z$ to the right hand sum

Multiplying $(2)$ by $e^{-z}$ yields $$ \frac{\mathrm{d}}{\mathrm{d}z}\left(f(z)\,e^{-z}\right)=\frac{1-e^{-z}}z\tag3 $$ Note that $$\newcommand{\Ei}{\operatorname{Ei}} \begin{align} \int_0^1\frac{1-e^{-x}}x\,\mathrm{d}x-\int_1^\infty\frac{e^{-x}}x\,\mathrm{d}x &=-\int_0^1\log(x)\,e^{-x}\,\mathrm{d}x-\int_1^\infty\log(x)\,e^{-x}\,\mathrm{d}x\tag{4a}\\ &=-\int_0^\infty\log(x)\,e^{-x}\,\mathrm{d}x\tag{4b}\\[6pt] &=\gamma\tag{4c}\\[6pt] \underbrace{\int_0^z\frac{1-e^{-x}}x\,\mathrm{d}x}_{\large f(z)\,e^{-z}}-\underbrace{\int_z^\infty\frac{e^{-x}}x\,\mathrm{d}x}_{\large-\Ei(-z)} &=\log(z)+\gamma\tag{4d} \end{align} $$ Explanation:
$\text{(4a)}$: integrate by parts
$\text{(4b)}$: combine the integrals
$\text{(4c)}$: apply this answer
$\text{(4d)}$: add $\int_1^z\frac1x\,\mathrm{d}x=\log(z)$

Solve for $f(z)$ in $\text{(4d)}$: $$ f(z)=e^z\left(\log(z)+\gamma-\Ei(-z)\right)\tag5 $$ Repeatedly integrating by parts yields $$ \begin{align} -e^z\Ei(-z) &=e^z\int_z^\infty\frac{e^{-x}}x\,\mathrm{d}x\tag{6a}\\ &=\frac1z-\frac1{z^2}+\frac2{z^3}-\frac6{z^4}+e^z\int_z^\infty\frac{4!}{x^5}\,e^{-x}\,\mathrm{d}x\tag{6b}\\ &=\frac1z-\frac1{z^2}+\frac2{z^3}-\frac6{z^4}+O\!\left(\frac1{z^5}\right)\tag{6c} \end{align} $$ Therefore, as $z\to\infty$, $$ f(z)=e^z\left(\log(z)+\gamma\right)+\frac1z-\frac1{z^2}+\frac2{z^3}-\frac6{z^4}+O\!\left(\frac1{z^5}\right)\tag7 $$