2
$\begingroup$

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 $-1$.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – user 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 $a_n \not \to 0$. $\endgroup$ – user 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! $\endgroup$ – user Mar 21 '18 at 6:57
3
$\begingroup$

HINT

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $a_n$? $\endgroup$ – user Mar 21 '18 at 6:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.