Limit involving a diverging series

Question

It is a known fact that the value of the sum $\Sigma \frac{1}{i}$ (when $i$ ranges over the integers) Diverges. But what is the solution of the following limit $$\lim_{n \to \infty} \frac{\Sigma_{i=0}^n 1/i}{n}$$

Answer 1

It is known that

$$ \frac{\ln (n+1)}{n}<\frac{\sum_{i=0}^{n} \frac{1}{i}}{n}<\frac{1+\ln n}{n}$$

Which follows from integration.

Thus $$0= \lim_{n \to \infty}\frac{\ln (n+1)}{n} \le \lim_{n \to \infty} \frac{\Sigma_{i=0}^n 1/i}{n} \le \lim_{n \to \infty}\frac{1+\ln n}{n} =\lim_{n \to \infty}\frac{1}{n}=0 (\because \text{L'Hospital})$$

Answer 2

By the Cauchy-Schwarz inequality $$ \sum_{n=1}^{N}\frac{1}{n}\leq \sqrt{N\cdot\sum_{n=1}^{N}\frac{1}{n^2}} \leq \sqrt{2N}$$ hence your limit is trivially zero.

Answer 3

If $a_n \to L, $ then $(a_1 + \cdots + a_n)/n \to L.$ Apply this with $a_n = 1/n,L=0.$