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Give a counter example to the statement:

$$ \lim n a_n=0 \Rightarrow \sum a_n \text{ converges}.$$

I've tried lots of series and sequencss but nothing seems to work.

Thanks in advance!

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up vote 6 down vote accepted

Take $a_n = \dfrac{1}{n\log(n)}, \forall n \geq 2$ for example.

Remark that $a_n \geq \int_{n}^{n+1}\dfrac{1}{x\log(x)}dx, \forall n \geq 2$, thus $$\sum_{n=2}^{\infty} a_n\geq \int_2^{\infty}\dfrac{1}{x\log(x)}dx = \log\log(x)|^{+\infty}_2 = +\infty$$

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There is a theorem not too hard to prove that says that if a series of positive decreasing terms $\sum_n a_n$ diverges, then $\sum_n a_n /s_n$ also diverges where $s_n = \sum_{j=1}^n a_j$ is the $n$th partial sum of the $a_j$. You can put $a_n = 1/n$ to get your example, by noting that the harmonic series diverges.

You can also use another not-too-hard theorem that a series of positive decreasing terms $\sum_n a_n$ converges if and only if $\sum_k 2^k a_{2^k}$ converges. If you let $a_n = n \log_2 n$ you see that the series diverges since the harmonic series $\sum_k 1/k$ diverges.

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+1. As a sidenote, the first theorem you mention can also be used to show a much more general (and interesting) statement (Abel): There exist no such function $\varphi(n)$ such that $a_n$ converges if and only if $\varphi(n)a_n \to 0$. The counterexample is $a_n = \frac{1}{\varphi(n)}$. – Winther Aug 25 '14 at 19:12

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