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Consider a sequence $a_n \ge 0$, $a_n \to +\infty$. Here, I'll say that a series $\sum{b_n}$ diverges if $\lim_{N\to \infty}{\sum_{n = 0}^N{b_m}} = \pm \infty$, and that doesn't converges if this limit doesn't exists. My aim is prove that $\sum_{n \ge 0}{(-1)^n a_n}$ can't neither converge (this seems to be pretty easy, and I guess I have shown it) nor diverge. So, my questions are

(a) Is that statement true?

(b) How can I exclude the divergence of the series (if possible)?

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How do you define diverges? Normally the series diverges just means the series does not converge, so a series must do one or the other. – Brian M. Scott Dec 15 '12 at 11:57
I say that a series $\sum_{n \ge 0}{a_n}$ diverges if $\lim_{N \to \infty}{s_N} = \pm \infty$, where $s_N:=\sum_{n = 0}^N{a_n}$, and that doesn't converge if that limit doesn't exists. – Jaques Dec 15 '12 at 12:02
I suspected that that was what you meant; you may want to bear in mind that it’s not the usual meaning of diverges. – Brian M. Scott Dec 15 '12 at 12:05
Thank you for your advice, but I don't think that my definition is so unusual. Of course, yours is Rudin's definition (and so, it could be considered a de facto standard), but several authors (like Giusti) define divergence and non-convergence as I did. – Jaques Dec 15 '12 at 12:16
The fact remains that it is not the usual meaning in English; I can’t speak for other languages. – Brian M. Scott Dec 15 '12 at 12:18
up vote 3 down vote accepted

Such a series can diverge (in your terms). Let $$a_n=\begin{cases}2^{\frac{n}2+1},&\text{if }n\text{ is even}\\\\2^{\frac{n-1}2},&\text{if }n\text{ is odd}\;,\end{cases}$$

so that $$\sum_{n\ge 0}(-1)^na_n=2-1+4-2+8-4+16-8+\ldots\;.$$

Show that $$\lim_{n\to\infty}\sum_{k=0}^n(-1)^ka_k=\infty\;.$$

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Necessary convergence condition says that if the series $\sum\limits_{n \ge 0}{u_n}$ converges, then $u_n \to{0}.$ Since $u_n=(-1)^n{a_n}\nrightarrow {0}$ then $\sum\limits_{n \ge 0}{(-1)^n{a_n}}$ cannot be convergent.

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