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?
 A: If you know that $1+\frac12 +\cdots+\frac1n\approx \ln n$ and $\frac{\ln n}n\to 0$, you are done.
More elementary, your limit is the Cesáro sum of the sequence $(\frac 1n)$, hence has the same limit $0$.
That is: 
$$\lim_{n\to\infty} a_n= a \qquad\Rightarrow\qquad \lim_{n\to\infty} \frac{a_1+\cdots+a_n}n= a$$
A: What you want to do here is find an upper bound for $1 + \frac{1}{2} + \frac{1}{3} + \ldots$.  That is, you want to replace every term in that series with a larger term in such a way that the sums are easier to do.  One clever way to do this (which I thought of by trying to modify the classic Oresme proof of divergence by exhibiting a lower bound) is to note that it's smaller than 
$$1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} \ldots.$$
The sum of the first $2^k-1$ terms of this sequence is $k$. Which should be enough for you to prove this limit directly.
A: Hint: compare the harmonic series with $\log n$, by comparing it with $\int_1^{n+1}\frac 1 x dx$. Then use l'Hopital's.
A: A slightly easier way out is prove that
$$1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n < 2\sqrt{n}$$ using induction.
Hence, you have that $$ 0 < \left(1 + \dfrac12 + \dfrac13 + \cdots + \dfrac1n\right) \dfrac1n < \dfrac2{\sqrt{n}}$$
Now use squeeze theorem to get what you want.
A: $$\sum_{k=1}^n\frac{1}{k}\sim \log n+\gamma$$
$$\lim_{n\to\infty} \frac{1}{n}\left(\log n+\gamma\right)=0$$
A: 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 $n \to \infty$.
