Estimating the quantity 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.
 A: 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.
A: 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 $$\begin{aligned}(1-e^{-H_n})^n &= \exp\left(n \log\left(1-\frac{e^{-\gamma}}{n} + o\left(\frac{1}{n} \right)\right)\right)\\ &= \exp\left(n\left(-\frac{e^{-\gamma}}{n} + o\left(\frac{1}{n} \right)\right)\right)\\
&= \exp(-e^{-\gamma})(1+o(1))\\ &= \exp(-e^{-\gamma}) + o(1)\end{aligned}$$
A: 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 $$\ee^x=1+x(1+x/2+...)\le 1+x/(1-x/2)\le 1+2x~\text{ for }~x\in[0,1].$$
