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Is there a divergent sequence such that $\lim_{k\rightarrow\infty}(x_{k+1}-x_k)=0$?

Pedro
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3 Answers3

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We know that $1/k \to 0$ so pick $x_k$ such that $x_{k+1} - x_{k} = 1/k$.

So start with $x_1 = 0$ and then define $x_{k+1} = 1/k + x_{k}$ now you can see this diverges since $\sum_{k = 1}^\infty 1/k$ does.

albatross
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Let $f:(0,\infty)\to\mathbb R$ be any function such that $\lim\limits_{x\to\infty}f(x)=\infty$ and $\lim\limits_{x\to\infty}f'(x)=0$. Then $(f(1),f(2),f(3),\ldots)$ is an example. In particular, you could take $f(x)=\log(x)$ or $f(x)=\sqrt x$.

Jonas Meyer
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Define $x_k:=\log k\ (k \ge 1)$, and then $x_{k+1}-x_k=\log(1+\frac{1}{k})\to 0$ as $k\to \infty$

Coiacy
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  • If only...and only if...it had not already been mentioned in the comments and in another answer. – Julien May 28 '13 at 01:04
  • @julien Sorry about that, and I found it's equivalent to the first one as $\sum_{i=1}^{n}\frac{1}{n} \sim \log n$ – Coiacy May 28 '13 at 01:06
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    No problem, I was only joking to quote your profile. – Julien May 28 '13 at 01:07
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    @Coiacy: In any case, you describe a different way to see that it is an example. Regarding the last part of your comment, this property is not preserved under asymptotic equivalence. For example $\log n +(-1)^n\sim\log(n)$. – Jonas Meyer May 28 '13 at 01:11
  • @Jonas Meyer Yeah, I agree and thanks. Actually you provide a more general answer. Well, next time before I post an answer, I'd better look at the posted answers carefully. – Coiacy May 28 '13 at 01:19