# Tricky complex series

gustav1139

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

## 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:

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.

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.

Staff Emeritus
Homework Helper
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.

Homework Helper
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...

gustav1139
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...

gustav1139
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.