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

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    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 – 2017-01-16
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    Yes, the series is divergent but the sequence isn't, or is it? – 2017-01-16
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    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 – 2017-01-16
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    Related: [Pseudo-Cauchy sequence](http://math.stackexchange.com/q/1237655) – 2017-01-16

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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.

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    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? – 2017-01-16
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    $a_{n}=H_{n}$, and as posted in the link, $H_{n}$ is divergent so $a_{n}$ is divergent. – 2017-01-16
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    @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 – 2017-01-16
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    @user406473 Do you understand? – 2017-01-16
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    but if I write my sequence now as Hn = 1/n, its konvergent. Sorry I guess I am really dumb right now lol. – 2017-01-16
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    @user406473 $H_{n}=\sum_{i=1}^{n} \frac{1}{i} \neq \frac{1}{n}$ – 2017-01-16
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    but that's a series.. – 2017-01-16
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    And as per my comment above, all that a series is is a sequence of partial sums – 2017-01-16
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    @user406473 No, a series can be a sequence. For example, let $a_{n}=\sum\limits_{i=1}^{n}i$. Then $a_{1}=1, a_{2}=3, a_{3}=6 \dots$ that is a sequence – 2017-01-16
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    is every series a sequence? – 2017-01-16
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    I'll say it a third time, a series is a sequence of partial sums. This is true for every series, yes, every series is a sequence of partial sums, i said it a fourth time. – 2017-01-16
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    @user406473 Every series is a sequence as long as it has some corresponding natural number to it. Every number can be a sequence as well. – 2017-01-16
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    alright thanks to both of you I think I understand it now :) – 2017-01-16
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Take $a_n=\ln n$. Then $$\lim_{n\to\infty}\Big[\ln(n+1)-\ln(n)\Big]=\lim_{n\to\infty}\ln\frac{n+1}{n}=\ln\left[\lim_{n\to\infty}\frac{n+1}{n}\right]=\ln 1=0$$