How to show that $\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i}=0 $? Show that 
$$\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =0 $$
I've proved that this sequence converges (it is bounded and decreasing). NOW, I need to find a sequence that is bigger than this one and goes to zero. Maybe something using geometric serie of 1/2
Thanks in advance!
 A: We can approximate a finite sum with a definite integral (see here). We obtain that
$$
\log(n+1)=\int_1^{n+1}x^{-1}\mathrm dx\le\sum_{i=1}^n\frac1i\le1+\int_1^nx^{-1}\mathrm dx=1+\log n.
$$
Now we need to show that
$$
\lim_{n\to\infty}\frac{\log n}n=0.
$$
This can be done by using l'Hôpital's rule (a more general statement is proved here).
A: In the same direction as user121270 and  coolydudey60 in their comments $$ \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =\frac{H_n}{n}$$ and for large values of $n$ $$\frac{H_n}{n}=\frac{\gamma +\log \left(n\right)}{n}+\frac{1}{2 n^2}-\frac{1}{12
   n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$ and then what V.C. proposed in his answer $\frac{1+\log(n)}{n}$ seems to be very good.
A: Using Stolz-Cesaro theorem we get
$$\lim\limits_{n\to\infty} \frac{\sum_{i=1}^n \frac1i}n = 
\lim\limits_{n\to\infty} \frac{\frac1{n+1}}{(n+1)-n} = 
\lim\limits_{n\to\infty} \frac1{n+1} =
0.$$ 
You can consider this as a special case of a more general fact that if $\lim\limits_{n\to\infty} a_n=L$, then also
$$\lim\limits_{n\to\infty} \frac{a_1+\dots+a_n}n=L.$$
(If a sequence is convergent, then the arithmetic means of the first $n$ element converge to the same limit.)
See for example this question (and other questions shown there among linked questions):
Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means
A: $$\sum_{i=1}^n \frac{1}{i} = \sum_{1\le i\le\sqrt{n}} \frac{1}{i} + \sum_{\sqrt{n}<i\le n} \frac{1}{i} \le \sum_{1\le i\le\sqrt{n}} 1 + \sum_{\sqrt{n}<i\le n} \frac{1}{\sqrt{n}} \le \sqrt{n} + \sqrt{n} = 2\sqrt{n}.$$
A: I just write an alternate approach, which may be useful:
Consider $\frac{1}{n}\sum_{i=1}^n\frac{1}{i}=\ln(a_n)$ then:
$$1\leq a_n=[e^{\frac{1}{1}}e^{\frac{1}{2}}...e^{\frac{1}{n}}]^{\frac{1}{n}}\leq \frac{e^{\frac{1}{1}}+e^{\frac{1}{2}}+...+e^{\frac{1}{n}}}{n}$$
Since $\lim_{n\rightarrow\infty}e^{\frac{1}{n}}=1$, the right-hand-side go to $1$ as $n$ approach $\infty$ (Cesaro mean), then $\lim_{n\rightarrow\infty} a_n=1$, which implies your limit is $\ln(1)=0$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With Stolz-Cesaro Theorem:

\begin{align}
\color{#66f}{\large\lim_{n\ \to\ \infty}{1 \over n}\sum_{i\ =\ 1}^{n}{1 \over i}}
=\lim_{n\ \to\ \infty}
{\sum_{i\ =\ 1}^{n + 1}1/i - \sum_{i\ =\ 1}^{n}1/i \over \pars{n + 1} - n}
=\lim_{n\ \to\ \infty}{1/\pars{n + 1} \over 1}=\lim_{n\ \to\ \infty}{1 \over n + 1}= \color{#66f}{\large 0}
\end{align}
A: It's much easier to prove that a Harmonic series is $O(\log n)$ and the result follows immediately
A: Another more general way is the following theorem:
Let's $a_n\ge0$ and $\sum_{n=1}^{\infty}a_n\lt \infty$ than
$$
\lim_{N\rightarrow \infty}\frac{\sum_{n=1}^{N}na_n}{N}=0.
$$
In this you just need to put $a_n=\frac{1}{n^2}$.
