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Does there exist a convergent sequence $\{a_n\}$ such that "$n d(a_{n+1} , a_n)$ is not bounded?

If not, how can i prove that $n d(a_{n+1} , a_n)$ is bounded? And is it right to conclude this;

This happens because domain of sequence is $\mathbb{N}$?

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up vote 6 down vote accepted

For example, let $a_{n}=(-1)^{n}\frac{1}{\sqrt{n}}$. Then $\lim_{n\rightarrow\infty}a_{n}=0$ and $nd(a_{n+1}-a_{n})=n(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}})$ come to infinity as $n$ come to infinity.

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Hint: Let $a_n=0$ if $n$ is odd. Let $a_n$ decrease slowly to $0$ over the evens.

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Let $a_n=1/2^k$ if $3^k\leqslant n\lt3^{k+1}$ for some integer $k\geqslant0$. The sequence $(a_n)_{n\geqslant1}$ is positive, nonincreasing, $(a_n)_{n\geqslant1}$ converges to $0$, but $n\cdot d(a_{n+1},a_n)=n\cdot (a_n-a_{n+1})$ is $0$ when $n$ is not a power of $3$, and is $3^k/2^k$ when $n=3^{k}$ for some integer $k\geqslant0$. In particular, the sequence of general term $n\cdot d(a_{n+1},a_n)$ is unbounded.

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This also answers the original, stronger form of the question before it was edited (once we replace $a_n$ by $-a_n$). – Erick Wong Oct 5 '12 at 16:05

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