My lectures notes say:

Given $$ \sum_{n=n_0}^{+\infty} (-1)^na_n$$ if $a_n=(-1)^n b_n $, with $b_n>0$ for every $n>n_0$ . If the series $$\sum_{n=n_0}^{+\infty}b_n$$ diverges and the the limit $$\lim_{n\rightarrow\infty}b_n$$ is not zero o does not exist, then by the necessary condition for convergence , the original series does not converge

I agree , based on the necessary condition for convergence, that the condition on the limit, implies the non-convergence of the series: $$\sum_{n=n_0}^{+\infty}b_n$$, which I may already know by other means, but how is that implies the non convergence of the original alternaiting series? And by saying non-convergence(instead of divergence), should I understand it has two possibilities: diverging or being indeterminate?


No series can converge without the general term tending to $0$. It doesn't matter whether the terms are positive or not. Here $(-1)^{n} a_n$ does not tend to $0$, so the series does not converge. This means that the partial sum sequence has no limit. It may oscillate or tend to $\pm \infty$.

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    $\begingroup$ @juancarlosvegaoliver $b_n \to 0$ iff $ a_n \to 0$ because $|a_n| =|b_n|$. So if you assume that $(-1)^{n} a_n $ converges to $0$ then you can conclude that $b_n \to 0$, But we are told that this is not the case. So we have arrived at a contradiction. This proves that $(-1)^{n} a_n $ cannot tend to $0$.. $\endgroup$ – Kavi Rama Murthy Jan 30 at 12:31
  • $\begingroup$ what I need to prove, precisely is that $(-1)^n a_n$ does not tend to zero, given the hypotheses, and not the other way around. You are starting with the thing we want to prove $\endgroup$ – J.C.VegaO Jan 30 at 12:32
  • $\begingroup$ @juancarlosvegaoliver I have edited my comment. $\endgroup$ – Kavi Rama Murthy Jan 30 at 12:36
  • $\begingroup$ ok the contradiction argument above is what I was looking for. One more thing, if it doesn't converge, I first thought it could oscillate or tend to $\pm\infty $. On second thought, I think it can only oscillate, because of the alternating signs, it can not diverge to $+\infty$ or $-\infty$, Am I right? $\endgroup$ – J.C.VegaO Jan 30 at 12:38
  • $\begingroup$ @juancarlosvegaoliver That is correct. It cannot tend to $\infty$ of $-\infty$. $\endgroup$ – Kavi Rama Murthy Jan 30 at 12:39

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