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Could any one find if the series:

$$\sum_{k=1}^{n} \frac{\sin(\sqrt{k})}{\sqrt{k}}$$

is divergent or convergent? I tried various techniques, but none of them worked (absolute convergence, Abel formula, inequalities, …). For example if we use Abel's formula, we don't know anything about $\sum_{k=1}^{n} \sin(\sqrt{k})$, so we can't conclude.

And now I don't know if this series needs cleverness or some advanced technique.

Daniel Fischer
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Abderrahim Abdou
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1 Answers1

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For every $m \in \mathbb{N}\setminus \{0\}$, we have $\sin \sqrt{k} \geqslant \frac{1}{2}$ when

$$\bigl(2m+\tfrac{1}{6}\bigr)\pi \leqslant \sqrt{k} \leqslant \bigl(2m+\tfrac{5}{6}\bigr)\pi.$$

Therefore

\begin{align} \sum_{(2m+1/6)^2\pi^2}^{(2m+5/6)^2\pi^2} \frac{\sin \sqrt{k}}{\sqrt{k}} &\geqslant \frac{1}{2} \sum_{(2m+1/6)^2\pi^2}^{(2m+5/6)^2\pi^2} \frac{1}{\sqrt{k}} \\ &> \sum_{(2m+1/6)^2\pi^2}^{(2m+5/6)^2\pi^2}\bigl(\sqrt{k+1} - \sqrt{k}\bigr) \\ &\approx \frac{2}{3}\pi. \end{align}

The sequence of partial sums is thus not a Cauchy sequence.

Daniel Fischer
  • 206,697
  • What about $\sum_{k} \frac{\sin(k^2)}{\sqrt{k}}$ ? It's interesting because the argument of the sine changes fast, as opposed to the original series. – Gabriel Romon Aug 19 '17 at 13:44
  • I don't know. $\sin k^{\alpha}$ is much easier to handle when $\alpha \leqslant 1$. No idea whether the oscillation of the partial sums of $\sum \sin (k^2)$ suffices to get convergence. – Daniel Fischer Aug 19 '17 at 13:54
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    The standard tools for dealing with $\sum_{n\geq 1}\frac{\sin(n^\alpha)}{n^\beta}$ are Van Der Corput's trick and Weyl's inequality. For instance $\sum_{n\geq 1}\frac{\sin(n^2)}{n}$ is convergent, but to study the convergence of $\sum_{n\geq 1}\frac{\sin(n^2)}{\sqrt{n}}$ is more difficult. This old question of mine is related: https://math.stackexchange.com/questions/215528/two-questions-about-weakly-convergent-series-related-to-sinn2-and-weyls-i – Jack D'Aurizio Aug 19 '17 at 14:03
  • @JackD'Aurizio Van der Corput lemma for oscillatory integrals or the lemma for uniform distribution mod $1$ ? – Gabriel Romon Aug 19 '17 at 14:16
  • @LeGrandDODOM: by Van Der Corput's trick I mean the trick for estimating the magnitude of $\sum_{k=1}^{n} e^{ik^2}$ by squaring. – Jack D'Aurizio Aug 19 '17 at 14:20