How to show $\lim_{n \rightarrow \infty}\frac{1}{n} \sum_{i=1}^{n}\frac{1}{i}=0$? How to show $\lim_{n \rightarrow \infty}\frac{1}{n} \sum_{i=1}^{n}\frac{1}{i}=0$?
I guess it is correct since $\lim_{x \rightarrow \infty}\frac{\log(x)}{x}=0.$ How to show it formally?
 A: Hint : Use Cauchy's first limit theorem.
Note that $\displaystyle \lim_{n\to \infty}\frac{1}{n}=0$.
A: Here's an elementary proof that $S=\sum_{i=1}^n\frac1i\in O(\sqrt{n})$: break the sum up into two pieces, $S=S_1+S_2$ where $S_1=\sum_{i=1}^{\sqrt{n}}\frac1i$ and $S_2=\sum_{i=\sqrt{n}}^n\frac1i$.  Now, $S_1$ is the sum of $\sqrt{n}$ terms each of size $\leq 1$, so $S_1\leq\sqrt{n}$; sumilarly, $S_2$ is the sum of $\lt n$ terms each of size $\leq\frac1{\sqrt{n}}$, so $S_2\lt\sqrt{n}$; summing, we have $S\leq 2\sqrt{n}$.  This gives $\frac1nS\leq \frac{2}{\sqrt{n}}$, and so the limit is clearly zero.
A: Hint: Use Cesaro's theorem noting that $a_n=1/n,\ \lim_{n\to \infty}a_n=0$. 
A: Here is a completely elementary proof
that only depends on
$\frac{k}{2^k}
\to 0
$
as
$k \to \infty
$.
If
$s_n
=\sum_{i=1}^n \frac1{i}
$,
$\begin{array}\\
s_{2n}
&=\sum_{i=1}^{2n} \frac1{i}\\
&=\sum_{i=1}^{n} \frac1{i}
\sum_{i=n+1}^{2n} \frac1{i}\\
&<\sum_{i=1}^{n} \frac1{i}
\sum_{i=n+1}^{2n} \frac1{n}\\
&=s_n+1\\
\end{array}
$.
Therefore
$s_{2^kn}
< s_n+k
$.
Also,
if
$2^kn 
\le m
< 2^{k+1}n
$,
$\begin{array}\\
s_m
&=\sum_{i=1}^m \frac1{i}\\
&=\sum_{i=1}^{2^kn} \frac1{i}
+\sum_{i=2^kn+1}^m \frac1{i}\\
&=s_{2^kn}+\sum_{i=2^kn+1}^m \frac1{i}\\
&\le s_{2^kn}+(m-2^kn)\frac1{2^kn}\\
&\le s_{n}+k+1\\
\end{array}
$
Therefore,
if
$2^kn 
\le m
< 2^{k+1}n
$,
dividing by $m$,
$\frac{s_m}{m}
\le \frac{s_{n}+k+1}{m}
\le \frac{s_{n}+k+1}{2^kn}
$
and this goes to zero
since
$\frac{k}{2^k}
\to 0
$
as
$k \to \infty
$.
A: It can be shown that
$$
\sum_{i=1}^{n}\frac{1}{i} = \log n + \gamma + o(1)
$$
as $n \to \infty$,
where $\gamma$ is the Euler constant; then
by the fact that $n^{-1}\log n \to 0$ as $n \to \infty$ we are done.
A: Use Stolz Theorem.
$$I=\lim_{n\to \infty} \frac{\frac{1}{n+1}}{(n+1)-n}=0$$
A: Since
$$
a_n=\frac1n\sum_{k=1}^n\frac1k\tag{1}
$$
is the average of a decreasing sequence, $a_n$ is a decreasing sequence.
Since each term is less than $\frac1n$ and there are $n$ terms, we have
$$
\sum_{k=n+1}^{2n}\frac1k\le1\tag{2}
$$
Therefore,
$$
\begin{align}
a_{2n}
&=\frac1{2n}\sum_{k=1}^{2n}\frac1k\\
&=\frac12\left(\frac1n\sum_{k=1}^n\frac1k\right)+\frac1{2n}\sum_{k=n+1}^{2n}\frac1k\\
&\le\frac12\left(a_n+\frac1n\right)\tag{3}
\end{align}
$$
Thus, setting $b_n=a_{2^n}$, $(3)$ becomes
$$
b_{n+1}\le\frac12\left(b_n+\frac1{2^n}\right)\tag{4}
$$
Multiplying $(4)$ by $2^{n+1}$ yields
$$
2^{n+1}b_{n+1}\le2^nb_n+1\tag{5}
$$
Since $b_0=1$, induction with $(5)$ gives
$$
a_{2^n}=b_n\le\frac{n+1}{2^n}\tag{6}
$$
Therefore, since $a_n$ is decreasing,
$$
\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{n+1}{2^n}=0\tag{7}
$$
