# Thread: Prove series divergence

1. ## Prove series divergence

$\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$

This is in the section covering alternating sequences. Leibniz's rule, conditional/absolute convergence, Dirichlet's test, and Abel's tests were all covered.

I don't know what to apply here, it seems like none of the tests I have learned are applicable. Most of the theorems in this section covered sufficient conditions for convergence, which I can't turn back on themselves RAA to prove that this diverges. Obviously the series does not absolutely converge, and the even terms are greater in absolute value than the odd terms, so it will keep increasing, however I think there's a problem with approaching it this way since it is essentially grouping terms of the infinite sequence within parenthesis, which is not strictly allowed.

2. ## Re: Prove series divergence

Originally Posted by process91
$\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$

This is in the section covering alternating sequences. Leibniz's rule, conditional/absolute convergence, Dirichlet's test, and Abel's tests were all covered.

I don't know what to apply here, it seems like none of the tests I have learned are applicable. Most of the theorems in this section covered sufficient conditions for convergence, which I can't turn back on themselves RAA to prove that this diverges. Obviously the series does not absolutely converge, and the even terms are greater in absolute value than the odd terms, so it will keep increasing, however I think there's a problem with approaching it this way since it is essentially grouping terms of the infinite sequence within parenthesis, which is not strictly allowed.
What is the [absolutely...] necessary condition for convergence of a series?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Prove series divergence

That the sequence most approach 0 as n increases without bound, but that is the case here.

$\displaystyle \lim_{n->\infty}\frac{(-1)^n}{\sqrt{n}+(-1)^n}=0$

Everything else which could prove divergence requires a positive series, so the only way I thought I could prove this would be to show that the sequence of partial sums $\displaystyle s_n=\sum_{k=2}^n \frac{(-1)^n}{\sqrt{n}+(-1)^n}$ is not bounded in some way. Of course, this summation does not have a nice summation formula.

The fact that the positive terms are greater (in absolute value) than their subsequent negative terms seems to indicate that $\displaystyle s_n<s_{n+1}$, but obviously this alone is not enough to prove divergence.

The book did not even cover a theorem which would justify my approach here, it only made this statement for positive sequences. It doesn't seem hard to justify this, though, since if the sequence of partial sums is not bounded then assuming it converged would be a contradiction.

4. ## Re: Prove series divergence

What I might suggest is write out some terms and then re-group terms and form a new series that doesn't alternate. Your standard tests should then apply.

5. ## Re: Prove series divergence

Originally Posted by process91
$\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$

This is in the section covering alternating sequences. Leibniz's rule, conditional/absolute convergence, Dirichlet's test, and Abel's tests were all covered.

I don't know what to apply here, it seems like none of the tests I have learned are applicable. Most of the theorems in this section covered sufficient conditions for convergence, which I can't turn back on themselves RAA to prove that this diverges. Obviously the series does not absolutely converge, and the even terms are greater in absolute value than the odd terms, so it will keep increasing, however I think there's a problem with approaching it this way since it is essentially grouping terms of the infinite sequence within parenthesis, which is not strictly allowed.
if you can show that every other partial sum does not converge, then....?

6. ## Re: Prove series divergence

(In reply to Danny...)

Sorry, do you mean something like

$\displaystyle \textstyle \frac{1}{\sqrt{2}+1} + \frac{-1}{\sqrt{3}+-1} + \frac{1}{\sqrt{4}+1} + \frac{-1}{\sqrt{5}+-1} ... = \left(\frac{1}{\sqrt{2}+1} + \frac{-1}{\sqrt{3}+-1}\right) + \left(\frac{1}{\sqrt{4}+1} + \frac{-1}{\sqrt{5}+-1}\right) ...$

I thought this was not a valid approach, since that would equivalently prove something like:

$\displaystyle 1 - 1 + 1 - 1 + 1 - 1 + ... = (1 - 1) + (1 - 1) + (1 - 1) + ... = 0$

7. ## Re: Prove series divergence

Originally Posted by Deveno
if you can show that every other partial sum does not converge, then....?
Yes, but subtracting two diviging series doesn't say anything!

8. ## Re: Prove series divergence

Originally Posted by Deveno
if you can show that every other partial sum does not converge, then....?
I'm not sure I understand what you mean, could you be a little more explicit?

9. ## Re: Prove series divergence

Originally Posted by process91
(In reply to Danny...)

Sorry, do you mean something like

$\displaystyle \textstyle \dfrac{1}{\sqrt{2}+1} + \dfrac{-1}{\sqrt{3}+-1} + \dfrac{1}{\sqrt{4}+1} + \dfrac{-1}{\sqrt{5}+-1} ... = \left(\dfrac{1}{\sqrt{2}+1} + \dfrac{-1}{\sqrt{3}+-1}\right) + \left(\dfrac{1}{\sqrt{4}+1} + \dfrac{-1}{\sqrt{5}+-1}\right) ...$

I thought this was not a valid approach, since that would equivalently prove something like:

$\displaystyle 1 - 1 + 1 - 1 + 1 - 1 + ... = (1 - 1) + (1 - 1) + (1 - 1) + ... = 0$
What I meant was

$\displaystyle \displaystyle \frac{1}{\sqrt{2}+1} + \frac{-1}{\sqrt{3}-1} + \frac{1}{\sqrt{4}+1} + \frac{-1}{\sqrt{5}-1} ...$

$\displaystyle \displaystyle = \left(\frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{4}+1} +\frac{1}{\sqrt{6}+1} \cdots \right) - \left(\frac{1}{\sqrt{3}-1} + \frac{1}{\sqrt{5}-1} +\frac{1}{\sqrt{7}-1} \cdots \right)$

$\displaystyle \displaystyle = \sum_{n=1}^{\infty} \frac{1}{\sqrt{2n}+1} - \frac{1}{\sqrt{2n+1}-1}$

10. ## Re: Prove series divergence

Originally Posted by Danny
What I meant was

$\displaystyle \displaystyle \frac{1}{\sqrt{2}+1} + \frac{-1}{\sqrt{3}-1} + \frac{1}{\sqrt{4}+1} + \frac{-1}{\sqrt{5}-1} ...$

$\displaystyle \displaystyle = \left(\frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{4}+1} +\frac{1}{\sqrt{6}+1} \cdots \right) - \left(\frac{1}{\sqrt{3}-1} + \frac{1}{\sqrt{5}-1} +\frac{1}{\sqrt{7}-1} \cdots \right)$

$\displaystyle \displaystyle = \sum_{n=1}^{\infty} \frac{1}{\sqrt{2n}+1} - \frac{1}{\sqrt{2n+1}-1}$
I see. The next chapter covers what we can and can't group, so I'm unfamiliar with that sort of approach. How about something like this, which is along the same lines:

Let $\displaystyle s_n$ be the partial sum. Then

$\displaystyle s_{2n+2} - s_{2n} = \frac{1}{\sqrt{2n+2}+1} - \frac{1}{\sqrt{2n+1}-1}$

Taking the derivative yields

$\displaystyle \frac{1}{\sqrt{2 n+1} (\sqrt{2 n+1}-1)^2}-\frac{1}{\sqrt{2 n+2} (\sqrt{2 n+2}+1)^2}$

Which, for $\displaystyle n \ge 1$, is strictly greater than 0, and so the difference between partial sums is strictly increasing. Therefore, the series must diverge.

We haven't covered a specific theorem to reference here, but the basic definition of divergence at the end should be enough to arrive at that conclusion.

Does this look valid?

11. ## Re: Prove series divergence

What you're essential showing is the terms in my series are not going to zero (nth term test). However, my question to you is how do you know this derivative is greater than zero for n greater or equation to 1?

EDIT. Actually the terms are going to zero. My bad.

12. ## Re: Prove series divergence

$\displaystyle 0<\sqrt{2n+1}-1<\sqrt{2n+2}+1$

$\displaystyle 0<(\sqrt{2n+1}-1)^2<(\sqrt{2n+2}+1)^2$

$\displaystyle 0<\sqrt{2n+1}(\sqrt{2n+1}-1)^2<\sqrt{2n+2}(\sqrt{2n+2}+1)^2$

$\displaystyle 0<\frac{1}{\sqrt{2n+2}(\sqrt{2n+2}+1)^2}<\frac{1}{ \sqrt{2n+1}(\sqrt{2n+1}-1)^2}$

$\displaystyle 0<\frac{1}{\sqrt{2n+1}(\sqrt{2n+1}-1)^2}-\frac{1}{\sqrt{2n+2}(\sqrt{2n+2}+1)^2}$

13. ## Re: Prove series divergence

Interesting problem. I believe that we can show that

$\displaystyle \dfrac{1}{\sqrt{2n}+1} - \dfrac{1}{\sqrt{2n+1}-1}} \le - \dfrac{1}{2n}$

which would give that the series diverges.

Any takers?

14. ## Re: Prove series divergence

Originally Posted by process91
I'm not sure I understand what you mean, could you be a little more explicit?
if a sequence of partial sums $\displaystyle \{s_n\}$ converges, every partial sum is "eventually" within some arbitrarily small distance (within epsilon, no matter how small epsilon is) of some real number (called the limit of the sequence).

this means that every other partial sum $\displaystyle s_2, s_4,\dots,s_n,s_{n+2},\dots$ must also "eventually" be within epsilon of that same limit.

but if the subsequence of every other partial sum does not converge, how can the whole sequence possibly converge?

yes, "grouping" is illegal in establishing what the limit actually IS, and re-arranging terms is even worse. but that's not we are doing here.

my claim is: if a sequence converges, then any (infinite) subsequence must converge, as well. so if a subsequence does NOT converge, then the original sequence does not converge.

it is important that we keep the terms of the partial sums in the same order, no "skipping" terms, or moving them around.

to address the counter-example: 1 + -1 + 1 + -1 + 1 + -1 +......

here, the alternating subsequences $\displaystyle \{s_{2k-1}\}$ and $\displaystyle \{s_{2k}\}$ both converge, the former to 1, and the latter to 0.

this tells us nothing, to actually know something, we would have to know that the original sequence converges (so that we can deduce the convergence of the subsequences, and we don't know this), or we would have to know that one of the subsequences diverged (to deduce the divergence of the "parent sequence"), and that isn't the case here.

15. ## Re: Prove series divergence

Originally Posted by Deveno
here, the alternating subsequences $\displaystyle \{s_{2k-1}\}$ and $\displaystyle \{s_{2k}\}$ both converge, the former to 1, and the latter to 0.
Actually, isn't that enough? Clearly the entire sequence cannot converge to any number then.

Suppose that it did converge to some number M. Then, for some N, we have that for all n>N, $\displaystyle |s_n-M|<1/10$. Additionally, for some P and Q, $\displaystyle n>P \implies |s_{2n-1}-1|<1/10$ and $\displaystyle n>Q \implies |s_{2n}|<1/10$. Herein lies a contradiction.

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