For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

kalakuja
Posts: 43
Joined: Sun Feb 06, 2011 10:15 am UTC

I've been going through this one assigment i got for a week now. I tried asking guidance from the prof at the uni but all i got was some mumble i couldn't make sense and i'm shy to return to him without having anything done. Series [imath]\sum_{n=1}^{\infty}{a_{n}b_{n}}[/imath] converges with every sequence [imath]b_n[/imath] , that [imath]lim_{n \rightarrow \infty}b_n=1[/imath]. Show that series [imath]\sum_{n=1}^{\infty}{a_{n}}[/imath] converges absolutely.

I've been trying to proof it with contradiction but i'm having trouble selecting the sequence b_n to fit every possible a_n. Any help would be greatly appreciated.

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

kalakuja wrote:Series [imath]\sum_{n=1}^{\infty}{a_{n}b_{n}}[/imath] converges with every sequence [imath]b_n[/imath] , that [imath]lim_{n \rightarrow \infty}\ b_n=1[/imath]. Show that series [imath]\sum_{n=1}^{\infty}{a_{n}}[/imath] converges absolutely.

Are you missing an absolute value around the [imath]b_n[/imath] in the limit? If you're not, I'm a little worried about your result.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Here's how I would do it...

Suppose [imath]\sum_n a_n[/imath] is conditionally convergent. Can you show how to choose your sequence {bn} so some partial sum of [imath]\sum_n a_n[/imath] is bigger than M? What if you have to make each bn close to 1?

Now, can you make this true for every M?

Try this, and let me know if you're still stuck. I can give a bigger hint.

jestingrabbit wrote:Are you missing an absolute value around the [imath]b_n[/imath] in the limit? If you're not, I'm a little worried about your result.

You shouldn't be.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

skeptical scientist wrote:
jestingrabbit wrote:Are you missing an absolute value around the [imath]b_n[/imath] in the limit? If you're not, I'm a little worried about your result.

You shouldn't be.

Right you are.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

kalakuja
Posts: 43
Joined: Sun Feb 06, 2011 10:15 am UTC

Well hmm i can pick the b_n so that it is (1) for the positive b_n and (-1) for the negative b_n until the partial sum is big enough and then continue with the 1s.
But the problem comes with the [imath]\sum{a_{n}b_{n}}[/imath]=M+rest, where the rest still converges into some value. So i can't make it to diverge skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

kalakuja wrote:Well hmm i can pick the b_n so that it is (1) for the positive b_n and (-1) for the negative b_n until the partial sum is big enough and then continue with the 1s.
But the problem comes with the [imath]\sum{a_{n}b_{n}}[/imath]=M+rest, where the rest still converges into some value. So i can't make it to diverge What if you have to make each bn close to 1, say, within epsilon. Can you still make the partial sum bigger than M?
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

skullturf
Posts: 556
Joined: Thu Dec 07, 2006 8:37 pm UTC
Location: Chicago
Contact:

This seems like a good problem and I, although a math PhD, blushingly admit I haven't found a general solution yet.

We want to show that if \sum a_n is conditionally convergent, then we can find b_n such that b_n approaches 1, and \sum a_n b_n is divergent.

It can be instructive to play around with specific examples. Suppose a_n = (-1)^n / n. Then one choice of b_n that works is

b_n = 1 + (-1)^n / ln(n).

This works because a_n b_n = a_n + 1 / (n*ln(n)), and the series \sum a_n b_n is the sum of the convergent series \sum a_n and the divergent series \sum [ 1 / (n*ln(n)) ].

I have some ideas about what to replace ln(n) with in the general case, but I can't quite figure out how to make it all work. Maybe if I give it time and/or sleep on it, it'll click.

EDIT: It's possible that I'm just not seeing a "trick" that's obvious once you see it, but this seems like a subtle problem.

Without loss of generality, say b_n = 1 + e_n = 1 + (sgn a_n) c_n. Then e_n and c_n approach 0.

Then \sum (a_n b_n) = \sum (a_n + e_n a_n) = \sum (a_n + c_n |a_n|) = \sum a_n + \sum (c_n |a_n|), which diverges if and only if \sum (c_n |a_n|) diverges.

We know \sum |a_n| diverges, so we're trying to show we can choose c_n approaching 0 so that \sum (c_n |a_n|) still diverges. Roughly speaking, this is saying that given any divergent series, we can find another divergent series whose terms are "smaller" in a specific sense. It seems believable to me, but presumably our choice of c_n would have to depend on a_n in a clever way, and I can't figure out how to make it work.

Ddanndt
Posts: 60
Joined: Fri Oct 02, 2009 4:18 pm UTC
Location: Paris

kalakuja wrote:
Series [imath]\sum_{n=1}^{\infty}{a_{n}b_{n}}[/imath] converges with every sequence [imath]b_n[/imath] , that [imath]lim_{n \rightarrow \infty}b_n=1[/imath]. Show that series [imath]\sum_{n=1}^{\infty}{a_{n}}[/imath] converges absolutely.

Does this mean : Show that if there is a sequence [imath]a_n[/imath] such that for every sequence [imath]b_n[/imath] converging to 1, [imath]\sum_{n=1}^{\infty}{a_{n}b_{n}}[/imath] converges, then [imath]\sum_{n=1}^{\infty}{a_{n}}[/imath] converges absolutely ??
God does not care about our mathematical difficulties — He integrates empirically.
—Albert Einstein

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

Ddanndt wrote:
kalakuja wrote:
Series [imath]\sum_{n=1}^{\infty}{a_{n}b_{n}}[/imath] converges with every sequence [imath]b_n[/imath] , that [imath]lim_{n \rightarrow \infty}b_n=1[/imath]. Show that series [imath]\sum_{n=1}^{\infty}{a_{n}}[/imath] converges absolutely.

Does this mean : Show that if there is a sequence [imath]a_n[/imath] such that for every sequence [imath]b_n[/imath] converging to 1, [imath]\sum_{n=1}^{\infty}{a_{n}b_{n}}[/imath] converges, then [imath]\sum_{n=1}^{\infty}{a_{n}}[/imath] converges absolutely ??

That's certainly the way that I took it. Its also true.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

kalakuja
Posts: 43
Joined: Sun Feb 06, 2011 10:15 am UTC

I've been playing with the skullturf's idea of the [imath]b_n=1+c_n[/imath], where c_n goes to 0.
Best i've tried was [imath]b_n=1+a_n-C[/imath], where [imath]\sum {a_n}=C[/imath]. Ending up with the sum term [imath]\sum {(a_n)^2}[/imath] but this doesn't always converge. edit: and uh the C is a sum not sequence so the b_n doesn't work at all to begin with ++\$_
Mo' Money
Posts: 2370
Joined: Thu Nov 01, 2007 4:06 am UTC

skullturf wrote:We know \sum |a_n| diverges, so we're trying to show we can choose c_n approaching 0 so that \sum (c_n |a_n|) still diverges. Roughly speaking, this is saying that given any divergent series, we can find another divergent series whose terms are "smaller" in a specific sense. It seems believable to me, but presumably our choice of c_n would have to depend on a_n in a clever way, and I can't figure out how to make it work.
It turns out that it doesn't have to be incredibly clever, only somewhat clever. Here's a (sort of) hint: Can you prove (without using the integral test) that [imath]\sum_{n=1}^\infty {1 \over n\log n}[/imath] diverges? What is it about 1/n and log n that makes this work?

Mindworm
Posts: 88
Joined: Wed Sep 22, 2010 4:06 pm UTC
Location: The dark place where the counterexamples live.

Skullturf's idea is indeed the solution (find a c_n for every diverging sum over a_n such that c_n goes to 0 and the sum over a_n*c_n still diverges).

Hint: No matter how far in the sequence you are, there is still some k such that the next k a_i add up to n, for every n.
The cake is a pie.

skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Here's another hint, a bit bigger than my previous ones, that might help.

Let ε>0, and let $b_n=\begin{cases}1+\epsilon &\text{ if } a_n \geq 0\\1-\epsilon &\text{ if } a_n < 0.\end{cases}$

Can you show that for all M, there is a natural number N such [imath]\sum_{n=1}^N a_nb_n >M[/imath]?

If you got this, and you're still stuck, open the spoiler:
Spoiler:
Let ε>0, and let $b_n=\begin{cases}1+\epsilon &\text{ if } a_n \geq 0\\1-\epsilon &\text{ if } a_n < 0.\end{cases}$

Can you show that for every natural number N1, there is a natural number N2 such that $\sum_{n=N_1+1}^{N_2} a_nb_n >1?$

I think the spoilered hint should be big enough that you can get it from there, but try it without opening the spoiler if you can.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

Here is a pretty sizeable hint I started writing before skep chimed in.

Spoiler:
Firstly, if [imath]b_n = 1[/imath] for all n, then you have that [imath]\sum a_n[/imath] converges, so that's something. It either converges conditionally or absolutely. Assume that the [imath]\sum a_n[/imath] is conditionally convergent. That implies that you have an increasing sequence i_n st [imath]\sum_{k= i_n +1}^{i_{n+1}} |a_k| \geq 2^n.[/imath]

Now, think about defining [imath]b_n[/imath] so that its constant for indexes between [imath]i_n+1[/imath] and [imath]i_{n+1}[/imath] where [imath]a_n[/imath] has the same sign.
Last edited by jestingrabbit on Fri May 06, 2011 8:40 pm UTC, edited 5 times in total.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

skullturf (paraphrased) wrote:Given any divergent positive series, can we find another divergent series whose terms are "smaller" in a specific sense? It seems believable to me, but presumably our choice of c_n would have to depend on a_n in a clever way, and I can't figure out how to make it work.

Yes. The easiest way to do this is:
Spoiler:
We'll define cn so that cn=an for a while, then cn=an/2 for a while, then cn=an/3 for a while, and so on. Since sum(an) is infinite, all the tails are infinite, so all tails are eventually bigger than 1, 2, 3, etc. In other words, we can make sure that the sum of all terms which are an/k is at least 1, before moving on to the next bigger value of k.

More formally:
Spoiler:
Since $$\sum_{n=0}^\infty a_n = \infty,$$ for any numbers N and k we can find some number M>N such that $$\sum_{n=N+1}^M a_n > k.$$ We define a sequence of natural numbers N0<N1<N2<N3<... recursively by letting N0=0, and for k≥1, letting Nk be minimal such that $$\sum_{n=N_{k-1}+1}^{N_k} a_n > k.$$ Now define cn=an/k, when Nk-1<n≤Nk. Then for all k, we have $$\sum_{n=N_{k-1}+1}^{N_k} c_n > 1,$$ which implies $$\sum_{n=1}^{\infty} c_n = \infty.$$ This shows that given any divergent positive series Σ an, there is another divergent series Σ cn, where cn≤an, such that cn/an tends to 0.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

Ddanndt
Posts: 60
Joined: Fri Oct 02, 2009 4:18 pm UTC
Location: Paris

Does anyone know how to prove/disprove this: " If the series [imath]S_n=\sum a_n b_n[/imath] and [imath]\sum a_n[/imath] are convergent then [imath]\sum b_n[/imath] is convergent" . If ever it's true then there could be a simpler way to solve the problem.
God does not care about our mathematical difficulties — He integrates empirically.
—Albert Einstein

Qaanol
The Cheshirest Catamount
Posts: 3069
Joined: Sat May 09, 2009 11:55 pm UTC

Ddanndt wrote:Does anyone know how to prove/disprove this: " If the series [imath]S_n=\sum a_n b_n[/imath] and [imath]\sum a_n[/imath] are convergent then [imath]\sum b_n[/imath] is convergent" . If ever it's true then there could be a simpler way to solve the problem.

That is both false and inapplicable. Even if it were true it could not be used to prove what is needed here. And it is not true, as shown by the counterexample [imath]a_n = n^{-2}[/imath], [imath]b_n = n^{-1}[/imath]. (Or even more simply, [imath]b_n = 1[/imath] for all n works.) An even more egregious example is [imath]a_n = n^{-3}[/imath], [imath]b_n = n[/imath].
Last edited by Qaanol on Fri May 06, 2011 11:07 pm UTC, edited 2 times in total.
wee free kings

skullturf
Posts: 556
Joined: Thu Dec 07, 2006 8:37 pm UTC
Location: Chicago
Contact:

Mindworm's hint was the trigger for me. Here's how I did it. (Attached PDF)

EDIT: I wrote my solution after reading Mindworm's hint, but before reading all the hints in the thread. My solution would probably look tidier if (as Skeptical Scientist did in one of his hints) I had chosen my denominators to simply be 1,2,3,... rather than sums of the a_n.

(Originally, playing around with n and log n, I thought the denominators "had" to be sums of the a_n. The "trick" I was missing is that if a series has an infinite sum, you can break it into a "chunk" > 1, another "chunk" > 2, and so on.)
Attachments
divergentseries.pdf
Last edited by skullturf on Sat May 07, 2011 2:05 pm UTC, edited 1 time in total.

skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Qaanol wrote:Even if it were true it could not be used to prove what is needed here.

It could, however, be used quite easily to prove that the statement we're asked to prove is false. I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

kalakuja
Posts: 43
Joined: Sun Feb 06, 2011 10:15 am UTC