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 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.

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: $$\lim_{n\to\infty} a_n= a \qquad\Rightarrow\qquad \lim_{n\to\infty} \frac{a_1+\cdots+a_n}n= a$$

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.