# if one sequence is convergent, so is the other one [duplicate]

Possible Duplicate:
If $\sum a_n b_n &lt;\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$

I have a question, can anyone help me?

Let $\{ a_i\} _{i=1} ^ { + \infty }$ be a sequence of positive real numbers such that for every sequence $\{ b_i\} _{i=1} ^ { + \infty }$ of positive real numbers satisfying the condition $\sum {b_n ^2} <+ \infty$ we have $\sum {a_n b_n} < + \infty$ . Prove that $\sum {a_n^2} < + \infty$.

It appeared in Iran's 3rd round Olympiad exam 2009, but I think it's a well-known result.

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– Nate Eldredge Feb 2 '12 at 18:19

## marked as duplicate by Davide Giraudo, Aryabhata, JavaMan, Nate Eldredge, DidFeb 2 '12 at 19:07

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.

## 1 Answer

For any $x \in \ell^2$, $\sum_n a_n x_n$ converges absolutely, and therefore converges. Now apply the Uniform Boundedness Principle to the sequence of "partial sum" functionals $x \to \sum_{n \le N} a_n x_n$, together with the Riesz-Fischer theorem that identifies bounded linear functionals on $\ell^2$ as members of $\ell^2$.

If you consider all this too functional-analytic for an Olympiad problem, a more elementary way would be this. If $\sum_n a_n^2 = \infty$, consider an increasing sequence $N_j$ such that $N_1 = 1$ and $c_j = \sum_{n=N_j}^{N_{j+1}-1} a_n^2 > 1$. Define a sequence $b_n$ by $b_n = a_n/(j \sqrt{c_j})$ for $N_j \le n < N_{j+1}$. Note that $\sum_{n=N_j}^{N_{j+1}-1} b_n^2 = 1/j^2$ and $\sum_{n=N_j}^{N_{j+1}-1} a_n b_n = \sqrt{c_j}/j > 1/j$, so $\sum_n b_n$ converges but $\sum_n a_n b)n$ diverges.

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