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


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

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  • $\begingroup$ What is your attempt? $\endgroup$ – Hanul Jeon Apr 25 '16 at 13:19
  • $\begingroup$ If you start with your partial proof maybe someone here will help you finish. $\endgroup$ – Ethan Bolker Apr 25 '16 at 13:19
  • $\begingroup$ Have a look at: math.stackexchange.com/questions/1757859/… $\endgroup$ – Martin Sleziak Apr 25 '16 at 15:07
  • $\begingroup$ 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.) $\endgroup$ – Martin Sleziak Apr 25 '16 at 15:10
  • $\begingroup$ Another question about the same limit: math.stackexchange.com/questions/221114/… $\endgroup$ – 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 $\lim_{k\rightarrow +\infty}{{\log(k)}\over k}$

  • $\begingroup$ Thank you for response. This proof is perfect. $\endgroup$ – wenbo Apr 25 '16 at 13:40

This can be done with the help of the Cesàro mean: Since $\lim_{m\to\infty} \frac 1 m = 0$, also $$\lim_{k\to\infty} \frac 1 k \sum_{m=1}^k \frac 1 m = 0.$$


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