I have tried to link with the series {$\sum \frac{1}{n}$} but it does not work.
How to give a example that $\lim\limits_{n\to\infty}n|a_{n}-a_{n+1}|=0$ and {$a_{n}$} divergent
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calculus
real-analysis
sequences-and-series
examples-counterexamples
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1Try $\dfrac{1}{n\log n}$. – 2017-01-15
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0$a_n=\dfrac{1}{n}$ works ! $$n|a_{n}-a_{n+1}|=n(\dfrac{1}{n}-\dfrac{1}{n+1})=\dfrac{n}{n(n+1)}=\dfrac{1}{n+1}$$ – 2017-01-15
1 Answers
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If we define $b_n = a_{n+1} - a_n$, then we can reframe the problem as follows: we're looking for a series $\sum_{n=1}^\infty b_n$ that fails to converge but satisfies $\lim_{n \to \infty}\frac{b_n}{1/n} = 0$.
In other words, we need a series for which $b_n$ is "smaller" than $1/n$, but still diverges. One working example is $b_n = \frac{1}{n\log n}$.