For $n \in \mathbb{N}$, denote by $H_n$ the $n$th Harmonic number. I want to prove the limit \begin{equation} \lim_{n \to \infty} \big(1 - \mathrm{e}^{-H_n}\big)^n = \mathrm{e}^{- \mathrm{e}^{-\gamma}}, \end{equation} where $\gamma$ is Euler's constant. By using the bound $\ln(n)+\gamma \leq H_n \leq \ln(n) + \gamma + 1/(2n)$, I can get \begin{equation} \big( 1 - \mathrm{e}^{-\gamma}/n \big)^n \leq \big(1 - \mathrm{e}^{-H_n}\big)^n \leq \mathrm{e}^{- \mathrm{e}^{-\gamma}} \mathrm{e}^{- \mathrm{e}^{-1/(2n)}}. \end{equation} But I did not quite get the right limit. The limit of the left-hand side is $\mathrm{e}^{- \mathrm{e}^{-\gamma}}$ while the limit of the right-hand side is $\mathrm{e}^{- \mathrm{e}^{-\gamma}} \mathrm{e}^{-1}$. Anyone sees how to prove the limit? Thanks very much.
4 Answers
Using the rule :
$$\text{If}\ \lim_{n\mapsto a} f(n)^{g(n)}=1^\infty$$
$$\text{Then}\ \lim_{n\mapsto a} f(n)^{g(n)}=\text{exp}\left({\lim_{ n\mapsto a}g(n)(f(n)-1)}\right)$$
Since we have
$$\lim_{n\mapsto \infty}\left(1-e^{-H_n}\right)^n=1^\infty$$
Then
\begin{align} \lim_{n\mapsto \infty}\left(1-e^{-H_n}\right)^n=\text{exp}\left(-\color{red}{\lim_{n\mapsto\infty}ne^{-H_n}}\right) \end{align}
Lets find the red limit
$$M=\lim_{n\mapsto\infty}ne^{-H_n}\\ \log(M)=\lim_{n\mapsto\infty}\left(\log(n)-H_n\right)=-\gamma\\ M=e^{-\gamma}$$
Thus
$$\lim_{n\mapsto \infty}\left(1-e^{-H_n}\right)^n=\text{exp}(-\color{red}{e^{-\gamma}})$$
The proof of $\lim_{n\mapsto\infty}\left(\log(n)-H_n\right)=-\gamma$ can be found here.
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Since $H_n = \log n + \gamma+o(1)$, we have $$\exp(-H_n) = \exp(-\log n)\exp(-\gamma)\exp(o(1)) = \frac{e^{-\gamma}}{n}(1+o(1)) =\frac{e^{-\gamma}}{n} + o\left(\frac{1}{n} \right) $$
Thus $$\begin{aligned}(1-e^{-H_n})^n &= \exp\left(n \log\left(1-\frac{e^{-\gamma}}{n} + o\left(\frac{1}{n} \right)\right)\right)\\ &= \exp\left(n\left(-\frac{e^{-\gamma}}{n} + o\left(\frac{1}{n} \right)\right)\right)\\ &= \exp(-e^{-\gamma})(1+o(1))\\ &= \exp(-e^{-\gamma}) + o(1)\end{aligned}$$
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Thanks a lot for your post and I can understand your solutions now. But could you take a look at what I wrote in the original post? Essentially I use the more explicit $1/(2n)$ to replace $o(1)$ and do the calculations (using $1 - \mathrm{e}^{-x}$ is an increasing function and then using the bound on $H_n$). But why I get an upper bound of the limit which is smaller that the lower bound of the limit? Here $\mathrm{e}^{-1} < 1$. Thanks a lot. – Richie Sep 15 '19 at 09:30
$$y_n=\big(1 - \mathrm{e}^{-H_n}\big)^n\implies \log(y_n)=n \log\big(1 - \mathrm{e}^{-H_n}\big) $$
$$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}-\frac{1}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ $${e}^{-H_n}=\frac{e^{-\gamma }}{n}-\frac{e^{-\gamma }}{2 n^2}+\frac{5 e^{-\gamma }}{24 n^3}+O\left(\frac{1}{n^4}\right)$$ $$1-{e}^{-H_n}=1-\frac{e^{-\gamma }}{n}+\frac{e^{-\gamma }}{2 n^2}-\frac{5 e^{-\gamma }}{24 n^3}+O\left(\frac{1}{n^4}\right)$$ $$\log\big(1 - \mathrm{e}^{-H_n}\big)=-\frac{e^{-\gamma }}{n}+\frac{e^{-2 \gamma } \left(e^{\gamma }-1\right)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(y_n)=-e^{-\gamma }+\frac{e^{-2 \gamma } \left(e^{\gamma }-1\right)}{2 n}-\frac{e^{-3 \gamma } \left(8-12 e^{\gamma }+5 e^{2 \gamma }\right)}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\color{blue}{y_n=e^{-e^{-\gamma }}\left(1+\frac{e^{-2 \gamma } \left(e^{\gamma }-1\right)}{2 n}+O\left(\frac{1}{n^2}\right)\right)}$$
Edit
Using more terms, let $t=e^{- \gamma}$ and writing $$y_n=e^{-e^{-\gamma }}\left(1+\sum_{k=1}^3 \frac{c_k}{n^k}+ O\left(\frac{1}{n^4}\right)\right)$$ the $c_k$ are $$\left( \begin{array}{cc} k & c_k \\ 1 & -\frac{1}{2} t(t-1) \\ 2 & \frac{1}{24} t \left(3 t^3-14 t^2+15 t-5\right) \\ 3 & -\frac{1}{48} t \left(t^5-11 t^4+35 t^3-42 t^2+21 t-3\right) \end{array} \right)$$
Evaluated numerically, $c_1 >0$,$c_2 <0$, $c_3 <0$. This makes the blue formula an upper bound and a rather good estimation formula. Using as it is for a few values on $n$, here are some results for the term inside brackets that is to say for $e^{e^{-\gamma }} \left(1-e^{-H_n}\right){}^n$
$$\left( \begin{array}{cccc} n & \text{approximation} & \text{exact} &\frac{\text{approximation} } {\text{exact} } \\ 10 & 1.012311137 & 1.012195609 & 1.000114136 \\ 20 & 1.006155568 & 1.006127633 & 1.000027765 \\ 30 & 1.004103712 & 1.004091443 & 1.000012220 \\ 40 & 1.003077784 & 1.003070924 & 1.000006839 \\ 50 & 1.002462227 & 1.002457853 & 1.000004363 \\ 60 & 1.002051856 & 1.002048826 & 1.000003024 \\ 70 & 1.001758734 & 1.001756512 & 1.000002218 \\ 80 & 1.001538892 & 1.001537193 & 1.000001697 \\ 90 & 1.001367904 & 1.001366563 & 1.000001339 \\ 100 & 1.001231114 & 1.001230028 & 1.000001084 \\ 110 & 1.001119194 & 1.001118298 & 1.000000895 \\ 120 & 1.001025928 & 1.001025175 & 1.000000752 \\ 130 & 1.000947011 & 1.000946369 & 1.000000641 \\ 140 & 1.000879367 & 1.000878814 & 1.000000552 \\ 150 & 1.000820742 & 1.000820261 & 1.000000481 \\ 160 & 1.000769446 & 1.000769023 & 1.000000422 \\ 170 & 1.000724185 & 1.000723810 & 1.000000374 \\ 180 & 1.000683952 & 1.000683618 & 1.000000334 \\ 190 & 1.000647955 & 1.000647655 & 1.000000299 \\ 200 & 1.000615557 & 1.000615287 & 1.000000270 \end{array} \right)$$
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Thanks very much for your post. By the way, for your second step $H_n = \ln(n) + \gamma + 1/(2n) - 1/(12n^2) + O(1/n^4)$, do you have an estimate of the big $O$ term? I want to know if the explicit bound $H_n \leq \ln(n) + \gamma + 1/(2n)$ is correct or only $H_n \leq \ln(n) + \gamma + 1/(2n-1)$ is correct. Numerically what I got is that the error term $1/(2n)$ is enough. – Richie Sep 15 '19 at 22:45
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@Richie. I think that my edit gives even better bounds for $y_n$ since $c_2<0$ and $c_3 <0$. For the expansion to high orders https://en.wikipedia.org/wiki/Harmonic_number which is $$H_n=\gamma+\log(n)+\frac{1}{2 n}-\frac{1}{12 n^2}+\frac{1}{120 n^4}-\frac{1}{252 n^6}+\frac{1}{240 n^8}-\frac{1}{132 n^{10}}+O\left(\frac{1}{n^{12}}\right)$$ – Claude Leibovici Sep 16 '19 at 02:15
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@Richie. Cocerning the inequality $H_n \lt \ln(n) + \gamma + 1/(2n)$ is the one to use. – Claude Leibovici Sep 16 '19 at 02:26
As to the upper bound being smaller than the lower bound: The formula $e^{-e^{a+b}}=e^{-e^a}e^{-e^b}$ is wrong.
You should get $$e^{-e^{a+b}} = e^{-e^{a}e^{b}}=e^{-e^a}e^{e^a(1-e^b)}.$$ With $a,b$ negative the exponent of the second factor is positive, thus the second factor larger than $1$.
Applied this gives the corrected second formula $\newcommand{\ee}{\mathrm{e}}$ \begin{align} \big( 1 - \ee^{-γ}/n \big)^n \leq \big(1 - \ee^{-H_n}\big)^n &\leq \ee^{- \ee^{-γ}} \ee^{\ee^{-γ}(1- \ee^{-1/(2n)})} \\ &\leq \ee^{- \ee^{-γ}} \ee^{\ee^{-γ}/(2n)} \leq \ee^{- \ee^{-γ}} \big(1+\ee^{-γ}/n\big). \end{align} The additional bounds use the alternating series bounds and the geometric bound of the exponential $$\ee^x=1+x(1+x/2+...)\le 1+x/(1-x/2)\le 1+2x~\text{ for }~x\in[0,1].$$
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Oh I see, I was careless about the factorization. Thanks a lot for your comments and post. – Richie Sep 15 '19 at 10:58