Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
4  
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
show 2 more comments

2 Answers

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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