Question is in the title. I would appreciate any help with this as I am a bit clueless.

  • $\begingroup$ Are you familiar with the fact that $\sum\limits_{k=1}^\infty \frac{1}{k}$ is divergent? Or rather, more accurately written, $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k}$ is divergent $\endgroup$ – JMoravitz Jan 16 '17 at 4:29
  • $\begingroup$ Yes, the series is divergent but the sequence isn't, or is it? $\endgroup$ – Lillia Jan 16 '17 at 4:34
  • $\begingroup$ And a series is a sequence of partial sums. S.C.B. already spelled out what I was trying to get at with my hints below $\endgroup$ – JMoravitz Jan 16 '17 at 4:35
  • $\begingroup$ Related: Pseudo-Cauchy sequence $\endgroup$ – Martin Sleziak Jan 16 '17 at 6:43

Note that if $a_{n}=H_{n}$ where $H_{n}$ denotes the $n$-th harmonic number, $$\lim_{n \to \infty} H_{n+1}-H_{n}=\lim_{n \to \infty} \frac{1}{n+1}=0$$

However, the Harmonic Series is divergent.

  • $\begingroup$ I am sorry, I don't quite understand as I have to find a sequence which has to be divergent but in your case it's konvergent isn't it? $\endgroup$ – Lillia Jan 16 '17 at 4:40
  • $\begingroup$ $a_{n}=H_{n}$, and as posted in the link, $H_{n}$ is divergent so $a_{n}$ is divergent. $\endgroup$ – S.C.B. Jan 16 '17 at 4:41
  • $\begingroup$ @user406473 you want a sequence (in this case $H_n$) where the sequence itself is divergent, but the related sequence of differences (in this case $H_{n+1}-H_n$) has limit equal to zero $\endgroup$ – JMoravitz Jan 16 '17 at 4:42
  • $\begingroup$ @user406473 Do you understand? $\endgroup$ – S.C.B. Jan 16 '17 at 4:45
  • 2
    $\begingroup$ @user406473 $H_{n}=\sum_{i=1}^{n} \frac{1}{i} \neq \frac{1}{n}$ $\endgroup$ – S.C.B. Jan 16 '17 at 4:46

Take $a_n=\ln n$. Then $$\lim_{n\to\infty}\Big[\ln(n+1)-\ln(n)\Big]=\lim_{n\to\infty}\ln\frac{n+1}{n}=\ln\left[\lim_{n\to\infty}\frac{n+1}{n}\right]=\ln 1=0$$


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