$\lim_{n\to \infty} 1/n(1+1/2+...+1/n)$ It seem to me that the limit

$$\lim_{n\to \infty} \frac{1}n\sum_{i=1}^n\frac{1}{i}=0,$$

but I can't see how to prove this to myself and I'm a bit wary of infinity times $0 $ situations.
Any tips would be appreciated.
 A: Elementary proof:
Let $\epsilon >0$, there is $N\in\mathbb N$ such as $N\epsilon >1$. 
Hence as  $\frac1{N+1}<\frac 1N < \epsilon$,
$$
\sum_{k=1}^{n} 1/k
= \sum_{k=1}^{N} 1/k + \sum_{k=N+1}^{n} 1/k
< N + \frac{n-N}{N+1} < N + n\epsilon
$$and for $n>N^2$, $N<n\epsilon$ so
$$
\sum_{k=1}^{n} 1/k < 2n\epsilon
\\
\frac1n\sum_{k=1}^{n} 1/k < 2\epsilon
\implies \limsup \frac1n\sum_{k=1}^{n} 1/k =0
$$
A: What are you allowed to use? The numerator is $O(\log n) < \log n +1$ (harmonic series). You can also set $\log n = t, n = e^t$. 
EDIT show the lower bound (easy): $\frac{1}{n} \to_n 0$. Now the upper bound: use $e^t \geq 1+t + t^2$, you get $\frac{t}{1+t+t^2}$, from this you can get the upper bound that also converges to $0$. 
A: Let us set $t_n = \frac{1}{n}\sum_{i=1}^n\frac{1}{i}$.
One idea is to use the following inequality: $$t_{2^k} \leq \frac{1}{2^k}\cdot(\frac{1}{1} + \frac{1}{1} + \underbrace{\frac{1}{2}+\ldots +\frac{1}{2}}_{\textrm{4 times}} + \ldots +
\underbrace{\frac{1}{k}+\ldots +\frac{1}{k}}_{2k\textrm{ times}} ) = \frac{2k}{2^k} = \frac{k}{2^{k-1}}.$$
Then note that $\lim_{k\to\infty} \frac{k}{2^{k-1}} = 0$. Using the fact that $t_n > 0$ and obtaining a similar inequality for general $t_n$ as the one above for $t_{2^k}$, you get the desired result that $\lim t_n = 0$.
A: The summation corresponds to $H_n$; for large values of $n$, its expansion is $$H_n=\gamma +\log(n) +\frac{1}{2 n}+O\left(\left(\frac{1}{n}\right)^2\right)$$ So, dividing this result by $n$ $$ \frac{1}n\sum_{i=1}^n\frac{1}{i}=\frac{H_n}{n}=\frac{\gamma +\log(n) }{n}+\frac{1}{2
   n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$and going to $\infty$, you find the $0$.
A: By Stolz–Cesàro theorem, your limit is equal to $$\lim \dfrac{\sum_{i=1}^{n+1}\dfrac{1}{i}  - \sum_{i=1}^{n}\dfrac{1}{i} }{n+1 -n} = \lim \dfrac{1}{n+1}$$
