# Relation of two complex series

1. Aug 23, 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 $\sum{|b_n|^2}<\infty$. Show that $\sum{|a_n|^2}<\infty$.

2. Relevant equations

3. The attempt at a solution

We know that $\lim{a_n}=0$, since if that were not the case then for $b_n=\frac{1}{n a_n}$, $\sum{|b_n|^2}<\infty$, but $\sum{a_n b_n}=\sum{\frac{1}{n}}$ diverges. Not sure that's helpful though.

My other thought was to try to prove the contrapositive, that given $\left\{a_n\right\}$ such that $\sum{|a_n|^2}$ diverges, we could find a $\left\{b_n\right\}$ such that $\sum{a_n b_n}$ diverges as well.

So we can find a sequence $\left\{n_k\right\}$ such that $\sum_{n_k+1}^{n_{k+1}}{|a_n|^2}>1$. Then of course, we'd like to pick b's in such a way that $\sum{a_n b_n}$, while they still converge in the square. But since we don't know how far apart the $n_k$ are, I can't figure out a way to do that. If $b_n=\left|\frac{\bar{a_n}}{n}\right|$, then the b's converge the way we want them to, but it's not clear that $\sum{a_n b_n}$ diverges. On the other hand, if we choose something that depends on $a_n$ in some way, which seems more promising in some ways, then it's not clear that the b's converge the way they're supposed to. For instance if $b_n=\left|\frac{\bar{a_n}}{c_n}\right|$, where $c_n=k$ when $k<n\leq k+1$, then the c's are growing pretty slowly (presumably) so I shouldn't think the b's would converge properly.

Last edited: Aug 23, 2012
2. Aug 23, 2012

### Bacle2

It seems like it may have to see with Cauchy-Schwarz:

We know if (bn,bn) < oo , then (an,bn)^2<oo . Let me try some more.

3. Aug 23, 2012

### vela

Staff Emeritus
This argument doesn't work for an=1 since $\sum{|b_n|^2}$ will diverge.

4. Aug 23, 2012

### gustav1139

.....$\sum{\frac{1}{n^2}}$ converges... doesn't it?

5. Aug 23, 2012

### vela

Staff Emeritus
D'oh! Never mind. ;)