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Let $a_n \to 0.$

Then $$\lim_{n \to \infty} (a_{n+1}-a_n)n $$ equals to ??

I have taken a few examples and got that the limit equals to zero. It seems that the limit is zero, but how to prove it in general ?

or my guess is wrong...

Please provide some hint. Thank you.

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  • $\begingroup$ Interestingly, this limit is true under the stronger assumption that $a_n$ is summable and monotonic. $\endgroup$ – Theo Bendit May 6 '19 at 4:12
  • $\begingroup$ @Zhanxiong with your example limit is zero. $\endgroup$ – Eklavya May 6 '19 at 4:14
  • $\begingroup$ @Zhanxiong Your counter example is apparently wrong, the limit is $0$ in your case. $\endgroup$ – StAKmod May 6 '19 at 4:14
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Even if your sequence is monotone, this may not be true. If you consider $a_{k}=\frac{1}{n}$ for $n^3\leq k<(n+1)^3 $, then $$\lim_{n\to\infty}(a_{n^3}-a_{n^3-1})(n^3-1)=\lim_{n\to\infty}\left(\frac{1}{n}-\frac{1}{n-1}\right)(n^3-1)=\lim_{n\to\infty}\frac{n^3-1}{n-n^2}=-\infty$$ So, $\{(a_{n^3}-a_{n^3-1})(n^3-1)\}$is a divergent subsequence of $\{(a_{n+1}-a_n)n\}$.

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No general conclusion here. For example let $a_n$ be the sequence $$0,1/2,0,1/4,0,1/6,\cdots$$ You have $$(a_{2k+1}-a_{2k})(2k)=-\frac{2k}{2k}=-1$$ and $$(a_{2k}-a_{2k-1})(2k-1)=\frac{2k-1}{2k}\to1$$

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Your guess is wrong. Consider $a_n=(-1)^n/\sqrt{n}, n\geq 1$ Then $$n(a_{n+1}-a_n)=(-1)^{n+1}n(1/\sqrt{n+1})+1/\sqrt{n}).$$

However, $n(1/\sqrt{n+1}+1/\sqrt{n})>2\sqrt{n}\to +\infty$, which excludes $(-1)^{n+1}n(1/\sqrt{n+2}+1/\sqrt{n+1})\to 0$.

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