Suppose you have the series: $$\sum_{n=1}^\infty (-1)^n\ln(n)$$ Does it converge or diverge? You cannot apply the alternating series test since $b_n$ is not decreasing. Similarly, you cannot apply the nth term test for divergence since the $\lim_{x \to \infty}(-1)^n$ is either $1$ or $-1$.

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    $\begingroup$ Recall that if $a_n \not \to 0$ we cannot have convergence. More rigorously $$\sum a_n \text{converges} \implies a_n \to 0$$ thus if $a_n \not \to 0$ it cannot converge. $\endgroup$ – gimusi Mar 21 '18 at 6:40
  • $\begingroup$ Do you mean if $b_n = \ln(n)$ and $\lim_{x \to \infty} b_n$ goes to $0$? You cannot take the limit of $a_n = (-1)^n * \ln(n)$. $\endgroup$ – Art Mar 21 '18 at 6:44
  • $\begingroup$ The relevant fact is that $a_n \not \to 0$. $\endgroup$ – gimusi Mar 21 '18 at 6:52
  • $\begingroup$ Oh ok . I think I understand. It could be $-\infty$ or $+\infty$, but definitely not $0$. $\endgroup$ – Art Mar 21 '18 at 6:54
  • $\begingroup$ Exactly it must be zero! $\endgroup$ – gimusi Mar 21 '18 at 6:57


Simply check for the necessary condition $a_n\to0$.

Recall indeed that

$$S_n=\sum^n a_k \implies a_n =S_n-S_{n-1}$$

then if

$$S_n \to L\implies a_n =S_n-S_{n-1}\to L-L=0$$

  • $\begingroup$ Under what test? AST? @gimusi $\endgroup$ – Art Mar 21 '18 at 6:39
  • $\begingroup$ It is the first basic check to do. Just observe that $$a_n =S_n-S_{n-1}$$ then suppose that $S_n\to L$, what about $a_n$? $\endgroup$ – gimusi Mar 21 '18 at 6:48

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