# 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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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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