# Convergence of series (Theoretical Question)

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1. Jun 20, 2016

1. The problem:

Ive been all afternoon struggling with this doubt. Its a bit more teoric than the rest of the exercices i did and i just cant seem to get around it so here it goes :
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Consider ∑ 1/an a convergent serie of positive terms.

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What´s the nature of Σ (-1)^n/(an^(2) +1)

2. Relevant equations
(Leibniz criterion) ? If ∑(-1)^(n)an is an alternated serie and the sucession(an) is decreasing and his limit→+00 an = 0 then i can say that ∑(-1)^(n)an is converging.

3. The attempt at a solution

My initial thought was:... a-ha! This is an alternate serie! Because of the (-1)^n ...

So if i prove that the serie 1/an is decreasing since i already know the lim an= 0 i can say that the serie:
(Leibniz criterion)

Σ (-1)^n/(an) IS CONVERGENT however i have two problems...

1- I dont know how to prove that 1/an is decreasing.

2- The serie that they ask me to study is different and even if i could prove that 1/an is decreasing i dont know if through algebric manipulation i could get to:

Σ (-1)^n/(an^(2) +1)

Im not even sure if im going the right way. Anyone has any clue? This is bothering me so much!atement, all variables and given/known data

Last edited by a moderator: Jun 20, 2016
2. Jun 20, 2016

### Staff: Mentor

Given that $\sum \frac 1 {a_n}$ is a convergent series with $a_n > 0$, can you say something about the series $\sum \frac 1 {a_n^2 + 1}$?
Are the terms of this series larger, smaller, or the same as, term by term, of the series $\sum \frac 1 {a_n}$?

3. Jun 20, 2016

The terms start in n= 1 and go up to +00

4. Jun 20, 2016

### Staff: Mentor

Fine. Did you read my post?

5. Jun 21, 2016

Oh sorry. yes i did, i notice that the terms are smaller. Are you saying that i can compare the serie
Σ 1/(an^(2) +1) with (1/an) using the criterion of geral comparison?

If an≤bn for n ≥ f then:

(1) -Σbn is convergent → Σan is convergent

(2) -Σan is divergent → Σbn is divergent

At a first glance it doesnt seem that useful cause since the terms are smaller i get that Σ 1/(an^(2) +1) is (an) and that 1/an is (bn).

So i guess now i can say that Σ 1/(an^(2) +1) is convergent?

However that serie that im trying to prove doesnt have just positive terms, Σ (-1)^n/(an^(2) +1) also has negative terms so i cant apply criterion of comparison. So i have no clue...

Last edited: Jun 21, 2016
6. Jun 21, 2016

### Staff: Mentor

Yes.
If you know that a series $\sum b_n$ converges, can you conclude anything about the alternating series $\sum (-1)^nb_n$? I.e., does it
a) converge,
b) diverge, or
c) is it impossible to determine?

7. Jun 21, 2016

I cant conclude nothing using the geral criterion of comparison cause it only applies to series with positive terms. The problem is that (-1)^n !!!

Im almost sure it converges but on the definition of this criterion it says it is explicitly for series with positive terms.
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I have this example in my head: Σ(1/n)

1/2,1/3,1/4,1/5 1/6...

If i join that (-1)^n it will be something like:

1/2,-1/3,1/4,-1/5 1/6... which also converges but using a example obviously in no valid explanation.
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I also noticed that 1/an looks like a dirichlet series however i cant also make that assumption! cause the "(an)" can be anything!!! If i could im almost sure i knew how to do it.

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is it valid to do something like this?

1) Σ(-1)^n/(an^(2) +1) =( -1)^nΣ(1/(an^(2) +1)

2) since i know 1/an is a serie of positives terms larger than Σ(1/(an^(2) +1) and 1/an is convergent , Σ(1/(an^(2) +1) is also convergent.

3)Since Σ(1/(an^(2) +1) is convergent, ( -1)^nΣ(1/(an^(2) +1) also is (just cause? ).

Last edited: Jun 21, 2016
8. Jun 21, 2016

### George Jones

Staff Emeritus
At the risk of gving too much away,

9. Jun 21, 2016

I know that if (bn) converges, (an) will also converge. So i know that Σ(1/(an^(2) +1) is convergent, However my problem is that the serie in question is Σ((-1)^n/(an^(2) +1) and since it has negative terms i cant apply that criterion. The same applies with bn. If i know bn is convergent what criterion should i use to consider bn(-1)^n also convergent?

What is on my mind is: If i know that a certain serie is convergent, the limit →+00 = 0 and it doesnt seem to change if just multiply it by (-1)^n

10. Jun 21, 2016

### George Jones

Staff Emeritus
Forget everything except that there is a series with terms $b_n$, and that the series $\sum b_n$ converges. What can you conclude?

11. Jun 21, 2016

I can conclude that bn limit →+00 = 0 and that an is convergent.

12. Jun 21, 2016

### Ray Vickson

Go right back to basics. If $b_n > 0$ and $S =\sum b_n$ is convergent. By definition, this means that
$$S(N) = \sum_{n=1}^N b_n$$
is a convergent sequence; that is, $S = \lim_{N \to \infty} S(N)$ exists and is finite. Note that
$$\sum_{n=1}^N b_n = \sum_{n \leq N \;\text{even}} b_n + \sum_{n \leq N \;\text{odd}} b_n .$$

13. Jun 21, 2016

Ok! i think i got it. Can anyone just say if this sounds like a reasonable answer?

They ask me to study nature of ∑(-1)^n/(an^(2) + 1) knowing that 1/an is convergent.

Step 1- I know that ∑(1/(an^(2) + 1) < 1/an .

Step 2- Since all the terms of the series in Step 1 are positive i can conclude by the criterion of geral comparison that since 1/an is convergent ∑(1/(an^(2) + 1) is also convergent.

Step 3- Consider bn = ∑(1/(an^(2) + 1) so
∑bn*(-1)^n = ∑(-1)^n/(an^(2) + 1)

According to leibniz criterion, if ∑ bn*(-1)^n is an alternate serie where the sucession bn is decreasing and convergent and its lim+00 = 0.(basically saying that bn must be convergent)
I can conclude that ∑ bn*(-1)^n = ∑(-1)^n/(an^(2) + 1) is convergent!

Last edited: Jun 21, 2016
14. Jun 21, 2016

### Ray Vickson

I can come up with a sequence in which the $b_n>0$ are not monotone non-increasing, but which go to zero quickly enough that the series $\sum b_n$ converges anyway. You could not apply your argument to such a case.

15. Jun 21, 2016

### Ray Vickson

Your Step 1 is wrong; you may have $1/(a_n^2 +1) < 1/a_n$ but there is no reason at all to assume you have $\sum 1/(a_n^2 +1) < 1/a_n$. When you write things like that you are just begging for marks to be taken off!

Anyway, as I said before, you cannot just apply an alternating-series argument, because the successive terms may not be decreasing each and every time.

16. Jun 21, 2016

Sorry, ive been reviewing the steps and what i really meant to write as Step 1 was :

Step 1- I know that (1/(an^(2) + 1) < 1/an . (i bugged the integral symbol on the previous post)

Step 2- Since Σ1/an and ∑(1/(an^(2) + 1) have just positive terms i can conclude by the criterion of geral comparison that since ∑1/an is convergent ∑(1/(an^(2) + 1) is also convergent.

Step 3- Consider bn = ∑(1/(an^(2) + 1) so
∑bn*(-1)^n = ∑(-1)^n/(an^(2) + 1)

According to leibniz criterion, if ∑ bn*(-1)^n is an alternate serie where the sucession bn is decreasing and convergent and its lim+00 = 0.(basically saying that bn must be convergent)
I can conclude that ∑ bn*(-1)^n = ∑(-1)^n/(an^(2) + 1) is convergent!

Note:Since bn is convergent cant i tell by D`Alembert criterion that if i do lim n->+00 bn+1/bn = <1 and so... doesnt this mean that bn is decreasing?

Last edited: Jun 21, 2016
17. Jun 21, 2016

### Ray Vickson

No. Consider an example like $b_n = |\sin(n)|/n^{1.1}$. We have $b_n \leq 1/n^{1.1}$, so $\sum b_n$ is dominated by the convergent $p$-series $\sum 1/n^{1.1}$, but the terms of $b_n$ do not form a non-decreasing sequence: we can have $b_{n+1} > b_n$ for infinitely many values of $n$. Nevertheless, we can establish the convergence of $\sum (-1)^n b_n$, essentially using the hints I supplied in post #12.

18. Jun 22, 2016

I´ve read and spent some time with post 12, but i just dont see how it does prove the convergence of ∑(-1)^nbn, could you provide me with any hint?

19. Jun 23, 2016

### Ray Vickson

Try writing out the first six or seven terms of both the finite sums $\sum_{n=1}^N b_n$ and $\sum_{n=1}^N (-1)^n b_n$, then look carefully at what you have. It is important to remember that for finite sums it is perfectly legitimate to re-order the terms and collect them in any convenient forms you want.

Last edited: Jun 23, 2016