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I know $ \sum_{n=2}^\infty \frac{1}{n*ln(n)} $ is divergent by the integral test or comparison test; however, I notice that it fails the Series Test For Divergence ($\lim_{n\to\infty}a_n \neq 0 \Rightarrow Divergence$). Can a series fail this test and still diverge?

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  • $\begingroup$ You answered your question yourself: You gave an example of a divergent series for which the terms being added converge to $0$. $\endgroup$ – Andrés E. Caicedo Oct 12 '14 at 18:22
  • $\begingroup$ Yeah, but I just want to know why it's a theorem if it doesn't hold in all situations. $\endgroup$ – jake.toString Oct 12 '14 at 18:23
  • $\begingroup$ The theorem says: If $A$ then $B$. It does not say $A$ and $B$ are equivalent. $\endgroup$ – Andrés E. Caicedo Oct 12 '14 at 18:25
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    $\begingroup$ (In fact, it is an almost useless test. In most cases of divergent series one actually encounters in practice, the terms being added still converge to $0$.) $\endgroup$ – Andrés E. Caicedo Oct 12 '14 at 18:31
  • $\begingroup$ I will buy that. That was pretty much where I was going next. Since this is looking to be the case, the test must be inconclusive in many cases. Thanks. $\endgroup$ – jake.toString Oct 12 '14 at 18:33
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Yes, decay of the summand as $ n \rightarrow \infty$ is a necessary, not sufficient condition for convergence of the series.

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a simple example is summation of $\frac{1}{n}$

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It goes in only one way. If $lim_{n\to \infty} a_n \neq 0$ then the series diverge. It's a minimal criterial, not a sufficient condition

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