Calculating a limit with Harmonic number I am trying to prove that 
$\lim_{n\to\infty} (\frac{\sum^n_{k=1} \frac{1}{k}} { \ln(n) })^{ \ln(n) } = e^γ$
where $γ$ the Euler-Mascheroni constant. We know that that:
$\lim_{n\to\infty} \frac{\sum^n_{k=1} \frac{1}{k}}{ \ln(n) } = 1$
By approximating the sum with integrals
$$
\ln(n+1)=\int_1^{n+1}\frac1x\mathrm dx\le\sum_{k=1}^n\frac1k\le 1+\int_1^n\frac1x\mathrm dx=1+\ln n
$$
since $1/k$ is decreasing for $k\ge 1$. We have that
$$
\frac{\ln(n+1)}{\ln n}=\frac{\ln n+\ln(1+\frac1n)}{\ln n}\to1
$$
as $n\to\infty$ and we obtain the result. But I can't calculate $\lim_{n\to\infty} (\frac{\sum^n_{k=1} \frac{1}{k}} { \ln(n) })^{ \ln(n) } = e^γ.$ Any help?
 A: Your estimates are a bit too course. Recall, $\gamma = \lim_{n\rightarrow\infty} H_n-\ln(n)$ and $e^x=\lim_{n\rightarrow\infty}(1+x/n)^n$. So in this case, 
$$\lim_{n\rightarrow\infty}\left(\frac{H_n}{\ln(n)}\right)^{\ln(n)}=\lim_{n\rightarrow\infty}\left( \frac{H_n-\ln(n)+\ln(n)}{\ln(n)} \right)^{\ln(n)}=\lim_{n\rightarrow\infty}\left(1+\frac{H_n-\ln(n)}{\ln(n)}\right)^{\ln(n)}.$$
Let $g_n:=H_n-\ln(n)$. Then the inside of the limit is equivalent to:
$$\exp[\ln(n)\ln(1+g_n/\ln(n))]=\exp[g_n+\ln(n)o(g_n/\ln(n))],$$
which gives the desired result since $g_n$ approaches a constant. 
A: We can write
$$\sum_{k=1}^n \frac1k =\log(n)+\gamma+O\left(\frac1n\right)\tag1$$
Dividing $(1)$ by $\log(n)$ reveals
$$\begin{align}
\left(\frac{\sum_{k=1}^n\frac1k}{\log(n)}\right)^{\log(n)}&=\underbrace{\left(1+\frac{\gamma}{\log(n)}\right)^{\log(n)}}_{\to \gamma}\underbrace{\left(1+\frac{O\left(\frac1{n}\right)}{\log(n)+\gamma}\right)^{\log(n)}}_{\to 1}
\end{align}$$
And we are done!
A: Want to show
$\lim_{n\to\infty} (\frac{\sum^n_{k=1} \frac{1}{k}} { \ln(n) })^{ \ln(n) } = e^γ
$.
Let
$f(n)
=(\frac{\sum^n_{k=1} \frac{1}{k}} { \ln(n) })^{ \ln(n) }
$.
$\begin{array}\\
g(n)
&=\ln(f(n))\\
&=\ln(n)(\ln(\sum^n_{k=1} \frac{1}{k})-\ln(\ln(n)))\\
&=\ln(n)(\ln(\ln(n)+\gamma+O(1/n))-\ln(\ln(n)))\\
&=\ln(n)(\ln\ln(n)+\ln(1+\gamma/\ln(n)+O(1/(n\ln(n)))-\ln(\ln(n)))\\
&=\ln(n)(\ln(1+\gamma/\ln(n)+O(1/(n\ln(n))))\\
&=\ln(n)(\gamma/\ln(n)+O(1/(\ln^2(n))))\\
&=\gamma+O(1/(\ln(n))))\\
&\to \gamma\\
\end{array}
$
