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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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    Thanks. I don’t mind learning different conventions; I just thought that it would be useful for you to be aware of the need to clarify.2012-12-15

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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.