# How to prove $\mathop {\lim }\limits_{k \to \infty } \frac{1}{k}\sum\limits_{m = 1}^k {\frac{1}{m}} = 0$ [closed]

where $k$ is a integer. Can you give a complete proof?

## closed as off-topic by gebruiker, user186170, Hans Lundmark, Crostul, DidApr 25 '16 at 13:26

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• What is your attempt? – Hanul Jeon Apr 25 '16 at 13:19
Hint: $\sum_{m=1}^{m=k}{1\over m}\leq \int_1^k{1\over x}dx=\log(k)$, thus the limit is inferior to $\lim_{k\rightarrow +\infty}{{\log(k)}\over k}$
This can be done with the help of the Cesàro mean: Since $\lim_{m\to\infty} \frac 1 m = 0$, also $$\lim_{k\to\infty} \frac 1 k \sum_{m=1}^k \frac 1 m = 0.$$