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This question already has an answer here:

Suppose $\{a_n\}_{n=1}^{\infty}$ is a sequence of real numbers (suppose also positive for simplicity) so that $$\sum_{n=0}^{\infty} a_n^2 = \infty$$ i.e. the sum diverges.

Can you necessarily find a square summable sequence $\{b_n\}_{n=1}^{\infty}\in\ell^2$ so that $\sum_{n=0}^{\infty} a_n b_n = \infty$ ?

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marked as duplicate by Nate Eldredge, user147263, user223391, BLAZE, user91500 Nov 20 '15 at 5:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Assume without loss of generality that $a_1\ne0$ and consider, for every $n\geqslant1$, $$ b_n=\frac{a_n}{A_n},\qquad A_n=\sum_{k=1}^na_k^2. $$ Then
$$ \sum_na_nb_n=\sum_n\frac{a_n^2}{A_n},$$ which diverges by a standard result (alternatively, see the argument in a comment below). On the other hand, $A_n^2\geqslant A_nA_{n-1}$ and $a_n^2=A_n-A_{n-1}$ hence, for every $n\geqslant2$, $$ b_n^2\leqslant\frac{a_n^2}{A_nA_{n-1}}=\frac1{A_{n-1}}-\frac1{A_n}. $$ Summing these yields, $$ \sum_{n\geqslant1}b_n^2\leqslant b_1^2+\sum_{n\geqslant2}\left(\frac1{A_{n-1}}-\frac1{A_n}\right)=b_1^2+\frac1{A_1}=\frac2{a_1^2}. $$ in particular, the series $\sum\limits_nb_n^2$ converges.

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  • $\begingroup$ Which standard result yields the divergence of $\sum_n a_n \, b_n$? $\endgroup$ – gerw Nov 19 '15 at 11:59
  • $\begingroup$ @gerw For every $i$ there exists $j>i$ such that $A_j\geqslant2A_i$ (because $A_n\to\infty$ when $n\to\infty$), hence $$\sum_{n=i+1}^j\frac{a_n^2}{A_n}\geqslant\sum_{n=i+1}^j\frac{a_n^2}{A_j}=1-{}{}\frac{A_i}{A_j}\geqslant\frac12.$$ $\endgroup$ – Did Nov 19 '15 at 15:01
  • $\begingroup$ Thank you for the explanation! $\endgroup$ – gerw Nov 19 '15 at 15:42
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If $a_n$ is unbounded, then it is easy. Otherwise wlog $|a_n| \le 1$. Choose integers $0=j_0 < j_1 < j_2 < \cdots$ such that $$ \sum_{n=j_{k-1}+1}^{j_k} a_n^2\in [2^k, 2^k+1) .$$ Now let $$ b_n = 2^{-2k/3} a_k \quad\text{if } j_{k-1} < n \le j_k .$$

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