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?
7
-
$\begingroup$ 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. CaicedoOct 12, 2014 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.toStringOct 12, 2014 at 18:23
-
$\begingroup$ The theorem says: If $A$ then $B$. It does not say $A$ and $B$ are equivalent. $\endgroup$– Andrés E. CaicedoOct 12, 2014 at 18:25
-
1$\begingroup$ (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. CaicedoOct 12, 2014 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.toStringOct 12, 2014 at 18:33
|
Show 2 more comments
3 Answers
$\begingroup$
$\endgroup$
Yes, decay of the summand as $ n \rightarrow \infty$ is a necessary, not sufficient condition for convergence of the series.
$\begingroup$
$\endgroup$
a simple example is summation of $\frac{1}{n}$
$\begingroup$
$\endgroup$
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
answered Oct 12, 2014 at 18:05