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

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?

• Are you interested in $z\to +\infty$?
– Gary
Commented Jun 30, 2021 at 8:00
• Is there some particular reason you're using an exponential generating function? Since $H_n$ is roughly $\log n$, the $n!$ denominator is going to utterly swamp the $H_n$ numerator, so the power series is entire and you won't be able to use something simple like singularity analysis to get an asymptotic. Commented Jun 30, 2021 at 8:08
• Yeah, I'm interested in $z\to+\infty$ @Gary Commented Jun 30, 2021 at 8:09
• You can use $$\log n +\gamma +\mathcal{O}\!\left( \frac{1}{n} \right)$$ to obtain $$\sum\limits_{n = 1}^\infty {\frac{{\log n}}{{n!}}z^n }+\gamma (e^z -1) + \mathcal{O}(1)\left(\operatorname{Ei}(z) - \gamma - \log z\right)=\sum\limits_{n=1}^\infty {\frac{\log n}{n!}z^n} +\gamma (e^z -1) +\mathcal{O}\!\left( \frac{e^z}{z} \right) = \sum\limits_{n = 1}^\infty {\frac{{\log n}}{{n!}}z^n } + \gamma e^z \left( {1 + \mathcal{O}\!\left( {\frac{1}{z}} \right)} \right).$$ It seems to me that $$\sum\limits_{n = 1}^\infty \frac{\log n}{n!}z^n \sim e^z \log z$$ but this needs to be proved.
– Gary
Commented Jun 30, 2021 at 8:17
• Someting like dlmf.nist.gov/2.10#iii may help you to complete the argument I proposed.
– Gary
Commented Jun 30, 2021 at 8:20

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).

• (+1) I worked a while on an answer but didn't see yours until I finished. I got the same result after a lot of work.
– robjohn
Commented Jun 30, 2021 at 13:45
• @robjohn: I think you should un-delete your answer - it is more detailed than this one. Commented Jun 30, 2021 at 13:48
• A least detailed answer would be to combine dlmf.nist.gov/5.4.E14, dlmf.nist.gov/6.6.E3 and dlmf.nist.gov/6.12.E1
– Gary
Commented Jun 30, 2021 at 14:01
• @metamorphy: I have undeleted my answer, but if it seems too close, I will delete it again.
– robjohn
Commented Jun 30, 2021 at 15:09

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 $$z\to\infty$$, $$f(z)=e^z\left(\log(z)+\gamma\right)+\frac1z-\frac1{z^2}+\frac2{z^3}-\frac6{z^4}+O\!\left(\frac1{z^5}\right)\tag7$$