computing the limit of $\frac{1}{n} \sum_{k=1}^{n}{\frac{1}{k}}$ Prove that the following limit is 0
$\lim\limits_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n}{\frac{1}{k}}$
Please I don't know how to do it :S. The harmonic series diverges thus it requires some special trick
 A: A crude bound on $\sum_1^n \frac{1}{k}$ is enough.  
For every positive integer $k$, we have $k\ge \sqrt{k}$. Thus
$$\frac{1}{k}\le \frac{2}{\sqrt{k}+\sqrt{k-1}}=2(\sqrt{k}-\sqrt{k-1}).$$
(We rationalized the denominator.)
Now sum, $k=1$ to $n$. Note the telescoping. We conclude that
$$\sum_1^n \frac{1}{k} \le 2\sqrt{n}.$$
Finally, divide by $n$ and let $n\to\infty$. 
A: Solution 1: Apply Stolz theorem (L'Hospital for sequences) We get to consider the limit $\lim \frac{1/n}{n-(n-1)}=0$, which is therefore the value of the original limit. 
Solution 2 Consider the integral $\int_{1/n}^{1}\frac{dx}{x}$ and compute Riemann sums for the uniform partition of size $1/n$. Divide by $n$. Apply L'Hospital.
Solution 3: Divide the terms in groups of size being powers of $3$. Majorize by the last term in each group. Do the same but using sizes being powers of $2$ and minorize by the first term in each group. This gives you a sandwich theorem.
Solution 4 We know that $\frac{1}{n}\rightarrow 0$. Therefore for every $e$ there is $N$ such that if $n>N$ then $$-e<\frac{1}{n}<e.$$
From this $$-e(n-(n-1))<(1+1/2+...+1/n)-(1+1/2+...+1/(n-1))<e(n-(n-1))$$
Sum for $n=N$ to $n=K$ we get $$-e(K-(N-1))<(1+1/2+...+1/K)-(1+1/2+...+1/(N-1))<e(K-(N-1))$$
Divide by $K$. 
$$-e(1-\frac{(N-1)}{K})<\frac{(1+1/2+...+1/K)}{K}-\frac{(1+1/2+...+1/(N-1))}{K}<e(1-\frac{(N-1)}{K})$$
Rearranging
$$-e(1-\frac{(N-1)}{K})+\frac{(1+1/2+...+1/(N-1))}{K}<\frac{(1+1/2+...+1/K)}{K}<e(1-\frac{(N-1)}{K})+\frac{(1+1/2+...+1/(N-1))}{K}$$ 
For some $K$ we must have that for all $k>K$
$$-e<\frac{1+1/2+...+1/K}{K}<e.$$
Therefore $\frac{1+1/2+...+1/n}{n}\rightarrow0$.
A: This is trivial by Cesaro summation,  since $1/n \to 0$.  
To elaborate  a bit,  it holds that if $u_n \to l < \infty$ then $\frac{1}{n}\sum_{k=1}^{n} u_k \to l$. 
I can provide a proof if needed (only resorting to the definition of limit, no integral) 
http://en.m.wikipedia.org/wiki/Ces%C3%A0ro_summation
A: This is a special case of the following theorem: if $\lim_{n \to \infty} a_n = a$, then $\lim_{n \to \infty} \frac{1}{n} \sum \limits_{k=1}^n a_k = a$.
Proof: For $\epsilon>0 $ choose $N$, such that for all $n>N$: $ |a_n-a|<\epsilon$.
Then we have $|\frac{1}{n} \sum \limits_{k=1}^n a_k - a|= |\frac{1}{n} \sum \limits_{k=1}^n (a_k - a)|<|\frac{1}{n} \sum \limits_{k=1}^N (a_k - a)|+\frac{n-N}{n}\epsilon <2\epsilon$ for sufficiently large $n$.
A: Let us prove that
$$
\sum_{k=1}^n\frac 1k\le1+\log n.
$$
Split the integral into parts
$$
\log n=\int_1^n x^{-1}\mathrm dx=\sum_{k=2}^n\int_{k-1}^{k} x^{-1}\mathrm dx.
$$
Observe that $k^{-1}\le x^{-1}$ for $x\in[k-1,k)$ since $f(x)=x^{-1}$ is a decreasing function. Hence, 
$$
\sum_{k=2}^n\int_{k-1}^kx^{-1}\mathrm dx\ge\sum_{k=2}^nk^{-1}\int_{k-1}^k\mathrm dx=\sum_{k=2}^nk^{-1}.
$$
Check this answer to conclude that $\frac1n\sum_{k=1}^nk^{-1}\to0$ as $n\to\infty$.
A: You can also get a rough upper bound like this: by the inequality between the arithmetic and quadratic means, we have
$$
\frac{1}{n}\sum_{k=1}^n \frac{1}{k} \leq \sqrt{\frac{1}{n}\sum_{k=1}^n \frac{1}{k^2}}.
$$
Now, it is known that the series $\sum_{k=1}^\infty \frac{1}{k^2}$ converges to some constant $C$ (which is also known, but we don't need the precise value here). So we have
$$
\frac{1}{n}\sum_{k=1}^n \frac{1}{k} < \sqrt{\frac{C}{n}}.
$$
The right hand side converges to $0$, therefore the original sequence also converges to $0$.
