Evaluating $\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^{N} \frac{1}{n}$ $$\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^{N} \frac{1}{n}$$
I'm working on a proof that, if I could show the previous expression evaluates to zero could be helpful, but I'm really doubtful about it. It must have been talked about before on this site I just haven't been able to find it. 
I know that the sum of the harmonic series diverges and the limit of $\lim_{N \to \infty} \frac{1}{N}=0$, so it comes down to a zero times infinity situation.
 A: if you want to do this directly and without integration, consider 
for an upper bound:
$\frac{1}{N} \sum_{n=1}^{N} \frac{1}{n} = \frac{1}{N}\cdot\big( \sum_{n=1}^{N} 1\cdot\frac{1}{n}\big)\leq \frac{1}{N}\cdot \big(N\big)^\frac{1}{2} \big(\sum_{n=1}^{N} \frac{1}{n^2}\big)^\frac{1}{2} = \frac{1}{\sqrt{N}}\cdot \big(\sum_{n=1}^{N} \frac{1}{n^2}\big)^\frac{1}{2} \leq \frac{C}{\sqrt{N}}$
for some constant $C$, where the first inequality is Cauchy-Schwarz.  
A: Hint:
You have a Riemann sum:
$$\sum_{n=2}^N\frac 1n<\int_1^N\frac{\mathrm dt}t=\ln(N),\quad\text{
so}\quad\displaystyle\frac{1}{N} \sum_{n=1}^{N} \frac{1}{n}\le\frac1N+\frac{\ln(N)}{N}.$$ 
A: HINT:
Find a simple upper bound of the integral of $\int_1^N \frac1x\,dx$ to show that $\sum_{n=1}^N\frac1n\le \log(N)+1$.
A: If you are aware of harmonic numbers
$$\sum_{n=1}^{N} \frac{1}{n}=H_N$$ For large $N$, we have
$$H_N=\gamma +\log \left({N}\right)+\frac{1}{2N}+O\left(\frac{1}{N^2}\right)$$
$$\frac{1}{N} \sum_{n=1}^{N} \frac{1}{n}=\frac{H_N}{N}=\frac \gamma  N+\frac {\log \left({N}\right)} N+\frac{1}{2N^2}+O\left(\frac{1}{N^3}\right)\sim \frac \gamma  N$$
Just by curiosity, compute for $N=10$; the exact value is $\frac{7381}{25200}\approx 0.292897$ while the truncated expansion would give $\frac{\gamma }{10}+\frac{\log (10)}{10}+\frac{1}{200}\approx 0.292980$.
