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I am trying to find a sequence such that $\lim_{n\to\infty} |a_n - a_{n+1}|=0$, but the $(a_n)$ diverges

I tried thinking something periodic might work like $\sin(2\pi n)$, but that is convergent sequence

Edit : Never mind Log[n] works great. Figured it out.

4 Answers 4

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Try the sequence of the partial sums of the harmonic series, i.e.

$a_n:=1+\frac{1}{2}+...+\frac{1}{n}$

Here, $\,|a_{n+1}-a_n|=\frac{1}{n+1}\xrightarrow [n\to\infty]{}0\,$ but, as we know, the series diverges and thus, its partial

sums sequence diverges as well.

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    Sheeesh!?? Don't you think it's weird we *all* knew what was going on and only you didn't? Perhaps if you said you don't know anything about series we *all* would have tried to make things easier to you...2012-10-07
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or $a_n=\sqrt{n}$ also works well.

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    @jak : $\sqrt{n}-\sqrt{n-1} = \frac{(\sqrt{n}-\sqrt{n-1})(\sqrt{n}+\sqrt{n-1})}{\sqrt{n}+\sqrt{n-1}}=\frac{1}{\sqrt{n}+\sqrt{n-1}} \to 0$ as $n\to \infty$2012-10-07
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Or consider the sequence $\bigl\{0,{1\over2},1,{2\over3},{1\over3},0,{1\over4},{2\over4},{3\over4},1,{4\over5},\ldots\bigr\}$.

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The sequence $a_n=\ln(n)$ has the desired property,

$\lim_{n\to \infty } |\ln(n)-\ln(n+1)| = \lim_{n\to \infty } \left|\ln\left(\frac{n}{n+1}\right)\right| = 0 \,.$

but, $\lim_{n\to \infty } \ln(n) =\infty\,. $