About converging series, homework
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About converging series, homework
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.
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
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Re: About converging series, homework
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.
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Re: About converging series, homework
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 {b_{n}} so some partial sum of [imath]\sum_n a_n[/imath] is bigger than M? What if you have to make each b_{n} 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.
You shouldn't be.
Suppose [imath]\sum_n a_n[/imath] is conditionally convergent. Can you show how to choose your sequence {b_{n}} so some partial sum of [imath]\sum_n a_n[/imath] is bigger than M? What if you have to make each b_{n} 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
"With math, all things are possible." —Rebecca Watson
 jestingrabbit
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Re: About converging series, homework
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.
Re: About converging series, homework
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
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
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Re: About converging series, homework
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 b_{n} 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
"With math, all things are possible." —Rebecca Watson
Re: About converging series, homework
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.
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.
Re: About converging series, homework
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
—Albert Einstein
 jestingrabbit
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Re: About converging series, homework
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.
Re: About converging series, homework
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_nC[/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
Best i've tried was [imath]b_n=1+a_nC[/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
Re: About converging series, homework
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?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.

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Re: About converging series, homework
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.
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.
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Re: About converging series, homework
Here's another hint, a bit bigger than my previous ones, that might help.
Let ε>0, and let [math]b_n=\begin{cases}1+\epsilon &\text{ if } a_n \geq 0\\1\epsilon &\text{ if } a_n < 0.\end{cases}[/math]
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:
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.
Let ε>0, and let [math]b_n=\begin{cases}1+\epsilon &\text{ if } a_n \geq 0\\1\epsilon &\text{ if } a_n < 0.\end{cases}[/math]
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:
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
"With math, all things are possible." —Rebecca Watson
 jestingrabbit
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Re: About converging series, homework
Here is a pretty sizeable hint I started writing before skep chimed in.
Spoiler:
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.
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Re: About converging series, homework
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:
More formally:
Spoiler:
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
"With math, all things are possible." —Rebecca Watson
Re: About converging series, homework
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
—Albert Einstein
Re: About converging series, homework
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.
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Re: About converging series, homework
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.)
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.)
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 divergentseries.pdf
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Last edited by skullturf on Sat May 07, 2011 2:05 pm UTC, edited 1 time in total.
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Re: About converging series, homework
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
"With math, all things are possible." —Rebecca Watson
Re: About converging series, homework
Thank you all for your help. I had to smash through all the spoilers to get it solved, but the process was worth it.
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