Homework Help: Convergence properties.

1. Nov 17, 2012

peripatein

Hello,

I am trying to prove/disprove the following claims:

(1) There exists a sequence a_n of positive number so that the series Ʃ a_n converges whereas the series Ʃ (a_n)^2 diverges.

I believe I managed to disprove that. Is it indeed false?

(2) There exists a sequence of real numbers a_n so that the series Ʃ a_n converges absolutely whereas the series Ʃ (a_n)^2 diverges.

I proved it. Is it indeed true?

(3) Let a_n and b_n be sequences of real numbers. Provided that lim n->∞ (b_n)/(a_n) = 1 and the series Ʃ a_n converges, then the series Ʃ b_n converges.

(4) Prove that if the series Ʃ a_n converges, then the series Ʃ (a_n)/(n^2) converges absolutely.

I am not sure how to prove that. I am also not allowed to explicitly make use of the Cauchy product theorem. May you please advise how to go about it?

2. Nov 17, 2012

hedipaldi

The first two are the same.

3. Nov 17, 2012

peripatein

I agree that they are eseentially the same.
How may I prove (3) and (4)? Any concrete suggestions?

4. Nov 17, 2012

hedipaldi

For (4) use Abel's theorem for convergence of general series in the form Ʃanbn

5. Nov 17, 2012

peripatein

I don't think we're allowed to use that. May you please formulate the theorem so I may be a better judge of that?

6. Nov 17, 2012

micromass

In (2), you proved that there was an absolutely convergent series such that $\sum a_n^2$ diverges? May I know what your example of such a series is??

For (3), write out what it means that the limit equals 1 and use a comparison test.

7. Nov 17, 2012

peripatein

Re (3), I have tried that approach but I am not allowed to express that quotient as lim b_n = lim a_n, since that would be tantamount to dividing by lim a_n (which is zero). So I am not sure how to proceed.

8. Nov 17, 2012

micromass

You know that $\lim_{n\rightarrow +\infty}\frac{b_n}{a_n}=1$. What does that mean by definition?

9. Nov 17, 2012

peripatein

That [|b_n - a_n|/|a_n|] < epsilon.

10. Nov 17, 2012

micromass

No. Please give the full definition.

11. Nov 17, 2012

peripatein

There exists an epsilon>0 so that for every n0>=n |(b_n/a_n) - 1| < epsilon.

12. Nov 17, 2012

micromass

No. You really should know that definition from the back of your head.

13. Nov 17, 2012

peripatein

Sorry, it's very late here. I meant to write: for every epsilon > 0 there exists n0 so that for every n>=n0 |(b_n - a_n) - 1| < epsilon. Better?

14. Nov 17, 2012

peripatein

b_n/a_n instead of b_n - a_n

15. Nov 17, 2012

micromass

OK. So let's take $\varepsilon=1$. Then there exists an $n_0$ so that for all $n>n_0$, we have

$$|b_n-a_n|<|a_n|$$

Now use the comparison test.

16. Nov 17, 2012

peripatein

Sigma |b_n - a_n| converges (based on comparison test) and sigma a_n converges notwithstanding, does that necessarily indicate that sigma b_n converges?

17. Nov 17, 2012

micromass

Yes. Try to figure out why.

18. Nov 17, 2012

peripatein

Is it simply because sigma b_n = sigma (b_n - a_n + a_n) = sum of two converging series, hence convergent itself?

19. Nov 17, 2012

hedipaldi

But you are not given that the series of an converges absolutely.

20. Nov 17, 2012

hedipaldi

Attached-see Abel's test for convergence.

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