Proving Convergence of Complex Series with Cauchy Condition | Homework Help

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Homework Statement



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

Homework Equations


The Attempt at a Solution



I tried going directly, at it, using the Cauchy condition on \sum{a_{n}b_{n}} to try to figure something out about the a_{n}, 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 \sum{|a_{n}|} diverges, then there exists a sequence \left\{b_{n}\right\} such that \sum{a_{n}b_{n}} 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 \lim{\frac{a_{n+1}b_{n+1}}{a_{n}b_{n}}}=r where r\in(0,1)), but that's unfortunately not a necessary condition for convergence.
So I'm stuck, and frustrated :(
 
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What you need to prove is that if \displaystyle \int_{0}^{\infty} a(x)b(x)\,dx has a finite value, so must \displaystyle \int_{0}^{\infty} |a(x)|\,dx. 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.
 
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.
 
gustav1139 said:
Then I tried to prove the contrapositive, that if \sum{|a_{n}|} diverges, then \sum{a_{n}b_{n}} 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. \sum{|a_{n}|} diverges but \sum{a_{n}b_{n}} doesn't.

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

That said, I don't have any comment about how to actually do the proof. Just wanted to point out your error.
 
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...
 
vela said:
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...
 
AlephZero said:
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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