# Finding a sequence a with $\lim_{ n\to ∞} (a_{n+1}-a_n)=0$ a:divergent [duplicate]

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

• 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 – JMoravitz Jan 16 '17 at 4:29
• Yes, the series is divergent but the sequence isn't, or is it? – Lillia Jan 16 '17 at 4:34
• 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 – JMoravitz Jan 16 '17 at 4:35
• Related: Pseudo-Cauchy sequence – Martin Sleziak Jan 16 '17 at 6:43

Note that if $a_{n}=H_{n}$ where $H_{n}$ denotes the $n$-th harmonic number, $$\lim_{n \to \infty} H_{n+1}-H_{n}=\lim_{n \to \infty} \frac{1}{n+1}=0$$
• $a_{n}=H_{n}$, and as posted in the link, $H_{n}$ is divergent so $a_{n}$ is divergent. – S.C.B. Jan 16 '17 at 4:41
• @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 – JMoravitz Jan 16 '17 at 4:42
• @user406473 $H_{n}=\sum_{i=1}^{n} \frac{1}{i} \neq \frac{1}{n}$ – S.C.B. Jan 16 '17 at 4:46
Take $a_n=\ln n$. Then $$\lim_{n\to\infty}\Big[\ln(n+1)-\ln(n)\Big]=\lim_{n\to\infty}\ln\frac{n+1}{n}=\ln\left[\lim_{n\to\infty}\frac{n+1}{n}\right]=\ln 1=0$$