I'm looking for a sequence of real numbers $x_n$ such that $\sum x_n^2 < + \infty $ but $\sqrt{n}\ x_n$ does not tend to zero at infinity. I tried with some trigonometric sequence in vain. Thanks in advance.
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Hi! To avoid down-votes and close-votes, please provide us some context for this question, such as: (a) Is this homework? (b) If so, what course are you taking? (c) What specific topic are you covering at the moment? (d) What do you know that you think might be connected? (e) If you're stuck, what are you stuck on? For example, do you know what to apply, but don't know how to apply it, or do you not know what to apply? Please put these facts in your original post, not as responses to this comment, as comments may be deleted without warning. – Brian Tung Dec 08 '23 at 19:33
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Also, you might consider putting in what trigonometric sequences you tried, and where they led (even if it doesn't seem to be fruitful). – Brian Tung Dec 08 '23 at 19:33
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Note that $\sum\frac{\sin n}n$ converge by Dirichlet test but both $\sum\frac{|\sin n|}n$ and $\sum\frac{(\sin n)^2}n$ diverge. So your attempt with trig series is a bit doomed because of the positivity of $(x_n)^2$ terms – zwim Dec 08 '23 at 22:59
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Your condition implies that you have infinitely many $|\sqrt n x_n|$ that are above some $l > 0$, therefore $|x_n|$ also above $l/\sqrt n$ infinitely often. That implies $x_n^2 \geq l/n$ infinitely often. So you want a sequence with this property. One could set
$$ x_n = \begin{cases} l/\sqrt n, \text{ if } n = k^4, k \in \mathbb{N^*} \\ 0, \text{ else.} \end{cases} $$ $\sum_n x_n = \sum_k l/k^2$ which is convergent, but of course $\sqrt n x_n = l$ for any $n = k^4$, so infinitely often.
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