# Convergance Tests For Infinite Series

1. Jan 3, 2010

### TheForumLord

1. The problem statement, all variables and given/known data
I need help in the next questions.
Prove or find counterexamples for the next propositions:

1. If the series [ Sigma (from n=1 to infinity) n*an ] converge then the series
[ Sigma (from n=1 to infinity) n*a(n+1) ] also converge.

2. If [Sigma (n=1 to infinity) of an ] is a positive converge series then the series
[ Sigma (n=1 to infinity) sqrt( an*a(n+1) ) ] converge.

3. IF the series [ Sigma from k=1 to infinity of a(2k-1) ] converge and the series the series [Sigma fron k=1 to infinity of a(2k) ] converge then the seriesl
[Sigma fron n=1 to infinity of an] also converge.

4. If lim_n->infinity_ n*an =0 then the series Sigma(an) converge.
2. Relevant equations
3. The attempt at a solution
I think that 1 is incorrect but I can't find any counterexample for it.
I'm almost sure that 2 and 3 are true, but 4 isn't...
Can't find any counterexamples for any of these propositions...
Help is NEEDED!
TNX everyone!

2. Jan 3, 2010

what have you tried?

3. Jan 3, 2010

### TheForumLord

Well...I am pretty bad at counterexample but:
in 1- I think that a series that include non-positive elements might do the work but I can't figure out how to construct it... Each try gives me a series that converges also for n*a(n+1)...
In 2-I am pretty sure this proposition is correct but I can't find an elegant way to prove it...
IN 3- It's pretty obvious that Sigma_an = Sigma_a(2k) +Sigma_a(2k-1) ...But is it realy the proper proof? Is it a right way?

In 4- I'm pretty sure it's incorrect but again, each try gives me an incorrect counterexample...

I'll be delighted if you'll be able to guide me around here and to tell me how should I think and see this question...

TNX a lot!

4. Jan 3, 2010

### Dick

For 4, an=1/n 'almost' works. At least n*an is constant. You need something that goes to zero just a LITTLE BIT faster than 1/n. Any ideas? Think of functions f(n) so that f(n) goes to infinity very slowly, then 1/(n*f(n)) might do it.

5. Jan 3, 2010

### TheForumLord

Hmmm...About 4- If we'll take f(n)=n^0.5=sqrt(n) it will do the work, isn't it?
Because lim_n*an= lim_1/sqrt(n) =0 but the series 1/n^(3/2) diverges (isn't it? )

Will you guide me in the other parts too please? :)

TNX a lot!

6. Jan 3, 2010

### Dick

No. sqrt(n) goes to infinity too fast. 1/n^(3/2) converges. It's a p series. You need something sloooower. Do you know a function that goes to infinity slower than ANY power? I haven't looked at the other ones in detail. I'll give you a general hint though. Think about what convergence means in terms of partial sums. Apply that to 3. Try using it to show that if n*a(n) converges then a(n) converges. Hint for 1: what can you say about the convergence of (n-1)*a(n)???

Last edited: Jan 3, 2010
7. Jan 3, 2010

### no_alone

Maybe
$$\frac{1}{n*ln(n)^2}$$
Maybe this will work?

8. Jan 4, 2010

### TheForumLord

I think we can take a_n=1/[n*ln(n) ] and it will work... becase 1/ln(n) ->0 but 1/nln(n) diverges...

It will work, isn't it?

no_alone-the counterexample you gave converges...Check it out

9. Jan 4, 2010

### vela

Staff Emeritus
Are you sure #4 is false? The limit seems to say a_n goes to zero faster than 1/n, which would suggest the series would converge. What if you apply the n-th-root test to $$|n a_n|$$?

10. Jan 4, 2010

### TheForumLord

Will it give me something? I think a_n=1/[n*ln(n) ] is a proper counterexample...
a_n diverges but n*an->0 ...

11. Jan 4, 2010

### vela

Staff Emeritus
Well, no, not if you found a counterexample. :)

12. Jan 4, 2010

### TheForumLord

Haha so my counterexample is good for this purpose indeed...

Can you guide me in the other parts of the question?

tnX!

13. Jan 4, 2010

### TheForumLord

Actually, I only need guidance to the second part and counterexample to the first part...
About the first part: I figured out that : Sigma_n*a(n+1) =Sigma_n*an - Sigma_an ...
So if we'll be able to find a series an that diverges to -infinity and n*an converges-we're done... But no matter what I try, I can't find any proper counterexample for this...

TNX a lot!

14. Jan 4, 2010

### Dick

You won't find a counterexample. Try applying Abel's convergence test to the product of the sequences n*a(n) and 1/n.

15. Jan 5, 2010

### TheForumLord

Wow I can't believe I missed it!
So by abel's test we get that Sigma_an converges...And from that we get
Sigma_n*a(n+1) converges...[because Sigma_n*a(n+1) =Sigma_n*an - Sigma_an]...
TNX a lot

BTW- So maybe I'm wrong and the third part of the question is not true? I'm pretty sure it is true because Sigma_an = Sigma_a(2k) +Sigma_a(2k-1) ... But maybe I'm wrong...

TNX a lot anyway! You were all very helpful!

16. Jan 5, 2010

### Dick

The third part is true and you wrote down the reason. Put finite summation to say N on both sides. Now let N->infinity. Partial sums. It's the easiest problem in the list.

17. Jan 5, 2010

### TheForumLord

Well...The weird thing is that 3 of the statements on the list are true...It doesn't fit the regular profile of our lecturer...But guess he chose to be nice :)

TNX a lot!