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Tricky complex series

  • Thread starter gustav1139
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  • #1
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Homework Statement



Suppose that [itex]\left\{a_{n}\right\}[/itex] is a sequence of complex numbers with the property that [itex]\sum{a_{n}b_{n}}[/itex] converges for
every complex sequence [itex]\left\{b_{n}\right\}[/itex] such that [itex]\lim{b_{n}}=0[/itex]. Prove that [itex]\sum{|a_{n}|}<\infty[/itex].

Homework Equations





The Attempt at a Solution



I tried going directly, at it, using the Cauchy condition on [itex]\sum{a_{n}b_{n}}[/itex] to try to figure something out about the [itex]a_{n}[/itex], but I got bogged down, and all the inequalities seemed to be pointing the wrong direction.
Then I tried to prove the contrapositive, that if [itex]\sum{|a_{n}|}[/itex] diverges, then there exists a sequence [itex]\left\{b_{n}\right\}[/itex] such that [itex]\sum{a_{n}b_{n}}[/itex] diverges as well. But I didn't get very far with that either.
I was able to prove it using the ratio test (i.e. if [itex]\lim{\frac{a_{n+1}b_{n+1}}{a_{n}b_{n}}}=r[/itex] where [itex]r\in(0,1)[/itex]), but that's unfortunately not a necessary condition for convergence.
So I'm stuck, and frustrated :(
 
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Answers and Replies

  • #2
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What you need to prove is that if [itex]\displaystyle \int_{0}^{\infty} a(x)b(x)\,dx[/itex] has a finite value, so must [itex]\displaystyle \int_{0}^{\infty} |a(x)|\,dx[/itex]. You can get around the absolute value assuming that a(x) resolves to a non negative real number for every x.

Try using the first mean value theorem for integration.
 
  • #3
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Thanks! I'll give that a try, but isn't it an issue that they're complex sequences, if I want to use the MVT?

Also, why can you assume a(x) is a nonnegative real? Or perhaps I'm misunderstanding what you mean by resolve.
 
  • #4
vela
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Then I tried to prove the contrapositive, that if [itex]\sum{|a_{n}|}[/itex] diverges, then [itex]\sum{a_{n}b_{n}}[/itex] must diverge as well. But I didn't get very far with that either.
You didn't get the contrapositive quite right. Consider, for example, an=1 and bn=1/n2. [itex]\sum{|a_{n}|}[/itex] diverges but [itex]\sum{a_{n}b_{n}}[/itex] doesn't.

You should have said: if [itex]\sum{|a_{n}|}[/itex] diverges, then there exists a sequence {bn} such that ##\displaystyle\lim_{n \to \infty} b_n = 0## and [itex]\sum{a_{n}b_{n}}[/itex] diverges.

That said, I don't have any comment about how to actually do the proof. Just wanted to point out your error.
 
  • #5
AlephZero
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You could make a proof by contradiction. Suppose ## \sum | a_i|## diverges. Therefore you can find a sequence of integers ##k_n## such that the partial sums ## \sum_{k_{n-1}+1}^{k_n} | a_i| > 1## for n = 1, 2, 3 ...

Now choose a series of b's that converge to 0, such that ##\sum a_ib_i## diverges.

Hint: you can choose b's such that every term ##a_i b_i## is real and non-negative...
 
  • #6
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You didn't get the contrapositive quite right.
Right you are. All fixed.
That (what you said) is however what I'd been working with so that's still no go....
 
  • #7
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You could make a proof by contradiction. Suppose ## \sum | a_i|## diverges. Therefore you can find a sequence of integers ##k_n## such that the partial sums ## \sum_{k_{n-1}+1}^{k_n} | a_i| > 1## for n = 1, 2, 3 ...

Now choose a series of b's that converge to 0, such that ##\sum a_ib_i## diverges.

Hint: you can choose b's such that every term ##a_i b_i## is real and non-negative...
Thanks! I think that broke it open. I had been restricting myself unnecessarily with what I could assume if ##\sum|a_i|## diverged.

I picked ##b_k = \frac{\overline{a_i}}{c_i |a_i|}## where the c's are the largest n such that i > k_n. Then the b's go to 0 since they do in modulus, and you can group the terms in the product series to give something bounded below by the harmonic series. boom.
 

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