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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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – gebruiker, Marco Cantarini, Hans Lundmark, Crostul, Did
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• What is your attempt? – Hanul Jeon Apr 25 '16 at 13:19
• If you start with your partial proof maybe someone here will help you finish. – Ethan Bolker Apr 25 '16 at 13:19
• Have a look at: math.stackexchange.com/questions/1757859/… – Martin Sleziak Apr 25 '16 at 15:07
• Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Apr 25 '16 at 15:10
• Another question about the same limit: math.stackexchange.com/questions/221114/… – Martin Sleziak Oct 16 '16 at 10:45

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 $$\lim_{k\to\infty} \frac 1 k \sum_{m=1}^k \frac 1 m = 0.$$