Find sequence whose sum of squares converges Prove that for every sequence $\{a_n\}$ such that $\sum\limits_{n=1}^{\infty}a_n^2=+\infty$ there exists sequence $\{b_n\}$ such that $\sum\limits_{n=1}^{\infty}b_n^2$ converges but $\sum\limits_{n=1}^{\infty}a_nb_n$ diverges.
How should I approach this problem?
 A: A UBP-based proof by contradiction works fine, but does not produce $b_n$. I'd rather give constructive proofs when possible and practical.
Divide the sequence $a_n$ into blocks $m_{k}\le n<m_{k+1}$. I think of each block as a vector $\vec v_k$ (they don't have to be of the same dimension). Since $\sum a_n^2=\infty$, there is enough material for arbitrarily  long vectors: let's make them so that $|\vec v_k|\ge 2^k$. 
Now we need another sequence of vectors $\vec w_k$ such that $\sum |\vec w_k|^2<\infty$ but $\sum \vec v_k\cdot \vec w_k=\infty$. So, they should be pretty short  but with large inner product: such as $\vec v_k\cdot \vec w_k=|\vec v_k|\,|\vec w_k|\ge 1$. Of course, $w_k=2^{-k}\vec v_k/|\vec v_k|$ does the job.
A: Consider the duality for the norm in $\ell^2$ (it holds even for sequences not in $\ell^2$, but we apply it here only to finite sequences):
$$\tag{1}
\sum_n|a_n|^2=\sup\left\{\left|\sum_na_nb_n\right|^2:\ \sum_n|b_n|^2=1\right\}.
$$
Now divide the sequence $\{a_n\}$ in chunks $E_k$ such that 
$$
\sum_{n\in E_k}a_n^2> 2^{3k/2}.
$$
So, applying $(1)$ in each block $E_k$, we find numbers $c_n$ with
$$
\sum_{n\in E_k}c_n^2=1,\ \ \ \sum_{n\in E_k}a_nc_n>2^{3k/2}.
$$
If we now let $b_n=c_n/2^{k/2}$, then
$$
\sum b_n^2=\sum_k\sum_{n\in E_k}\frac{c_n^2}{2^k}=\sum_k\frac1{2^k}=1,
$$
and
$$
\sum_na_nb_n=\sum_k\sum_{n\in E_k}\frac{a_nc_n}{2^{k/2}}>\sum_k\frac{2^{3k/2}}{2^{k/2}}=\sum_k2^k=+\infty.
$$
A: For any arbitrary positive termed series,$a_{n}$,we have three possibilities:
1.$a_{n}$ converges to a limit,$l<1$.
2.$a_{n}$ converges to a limit,$l>1$.
3.$a_{n}>1,\forall n \in\mathbb{N}$,diverges.
4.$a_{n}=1,\forall n\in \mathbb{N}$ 
Case 1:
If $l<1$$\Rightarrow 0<a_{n}<1$.
We know that $\infty<\sum a^2_{n}<\sum a_{n}\Rightarrow \sum a_{n}$ diverges.
Also,$\forall n\in\mathbb{N},\sum a^n_{n}$ converges since by the root test,
$\lim (a^n_{n})^{1/n}=a_{n}<1$.
Now,to prove the existence of $b_{n}$,it suffices to show that atleast one such $b_{n}$ exists.Let us define $b_{n}:=a^n_{n}$.As shown above,by the root test $\sum b^2_{n}$ converges.
Now,$\sum a_{n}b_{n}=\sum a_{n}a^n_{n}$
Also since we had $0<a_{n}<1$,therefore ,$\exists c\in [0,1]$,such that $c<a_{n}\Rightarrow r=c^{n}<a^{n}_{n}$.
Therefore,$r\sum a_{n}<\sum a_{n}a^n_{n}$
Now,since $\sum a_{n}$ diverges therefore $r\sum a_{n}$ diverges,and by the comparison test,$\sum a_{n}a^n_{n}$ diverges.
Case 4:
If $a_{n}=1,\forall n \in \mathbb{N}$,let $b_{n}:=1/n$,
then we know that $\sum a^2_{n}$ diverges,$b^2_{n}=\sum \frac{1}{n^2}$ converges,
and $\sum a_{n}b_{n}=\sum \frac{1}{n}$,diverges.
Case 2 and 3:
I'll work these out later.
