# Tricky complex series

1. Aug 22, 2012

### gustav1139

1. The problem statement, all variables and given/known data

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 $\sum{|a_{n}|}<\infty$.

2. Relevant equations

3. 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 :(

Last edited: Aug 22, 2012
2. Aug 22, 2012

### Millennial

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.

3. Aug 22, 2012

### gustav1139

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. Aug 22, 2012

### vela

Staff Emeritus
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.

5. Aug 22, 2012

### AlephZero

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. Aug 22, 2012

### gustav1139

Right you are. All fixed.
That (what you said) is however what I'd been working with so that's still no go....

7. Aug 22, 2012

### gustav1139

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.