Find $\lim_{n\to\infty} (1+\frac{1}{2}+...+\frac{1}{n})\frac{1}{n}$ 
Find the following limit:
  $$\lim_{n\to\infty} \left(1+\frac{1}{2}+...+\frac{1}{n}\right)\frac{1}{n}$$

My intuition says that this goes to zero, because $1/n$ goes much faster to zero than the harmonic series go to infinity, but how can I prove this?

Here is a completely elementary
and self-contained proof
(subject to accepting that
$\frac{k}{2^k} \to 0$
as $k \to \infty$).
Let
$a_n
= \frac{1}{n}\sum_{k=1}^n \frac1{k}
$.
Then
$na_n
= \sum_{k=1}^n \frac1{k}
$.
Therefore
$\frac1{n+1}
=(n+1)a_{n+1}-na_n
=n(a_{n+1}-a_n)+a_{n+1}
$
or
$a_{n+1}-a_n
=\frac1{n}(\frac1{n+1}-a_{n+1})
$.
Since,
for $n > 1$,
$a_n
\ge \frac1{n}(1+\frac12)
= \frac{3/2}{n}
$,
$a_{n+1}-a_n
<-\frac{1}{2n(n+1)}
$
so
$a_n$
is a decreasing sequence.
We also have
$\begin{array}\\
(2n)a_{2n}-na_n
&=\sum_{k=1}^{2n} \frac1{k}-\sum_{k=1}^n \frac1{k}\\
&=\sum_{k=n+1}^{2n} \frac1{k}\\
\text{so}\\
(2n)a_{2n}-na_n
&< 1\\
\text{and}\\
(2n)a_{2n}-na_n
&> \frac12\\
\end{array}
$
Therefore
$a_{2n}
< \frac1{2n}(1+na_n)
= \frac1{2n}+\frac12 a_n
$
or
$\begin{array}\\
a_n
&>2a_{2n}-\frac1{n}\\
&>2(2a_{4n}-\frac1{2n})-\frac1{n}\\
&=4a_{4n}-\frac1{n}-\frac1{n}\\
&=4a_{4n}-\frac{2}{n}\\
&=4(2a_{8n}-\frac1{4n})-\frac{2}{n}\\
&=8a_{8n}-\frac1{n}-\frac{2}{n}\\
&=8a_{8n}-\frac{3}{n}\\
\end{array}
$
By induction
we can show that
$a_n
> 2^ka_{2^kn}-\frac{k}{n}
$
or
$a_{2^kn}
<2^{-k}(a_n+\frac{k}{n})
$.
Since both
$\frac{k}{2^k}
\to 0
$
and
$\frac{a_n}{2^k}
\to 0
$
as
$k \to \infty$,
$a_{2^kn}
\to 0
$
as
$k \to \infty$.
Since
$a_n$
is decreasing and positive,
$a_n \to 0$
as $n \to \infty$.
