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if we look to those trigonometric Sum : $\sum\limits_{k=1}^\infty \frac{\sin(k)}{\sqrt{k}},\sum\limits_{k=1}^\infty \frac{\cos(k)}{\sqrt{k}}$ one can say they could be diverge since $\sin $ and $\cos $ functions are bounded and lie in $[-1;1]$ , one can get two divergent sum if we take $ \sin(k)$ is $ 1$ for $ k \to \infty $ so we have $\sum_{k\geq1} \frac{1}{\sqrt{k}}$ which is diverge the same idea with $\cos $ function, Now i want if there is any theorem with proof show why exactly the titled sum are conveges ? and what is its geometric interpretation ?

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closed as off-topic by Zacky, Cesareo, Eevee Trainer, Song, Leucippus Mar 15 at 4:17

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  • $\begingroup$ Because of Dirichlet's test and this. $\endgroup$ – rtybase Mar 14 at 22:34
  • $\begingroup$ For $x \mathbb{R}, \not \in 2\pi \mathbb{Z}$ then $\sum_{k=1}^\infty e^{ik x} k^{-s} = \sum_{k=1}^\infty (\sum_{n=1}^k e^{inx}) (k^{-s}-(k+1)^{-s}) = \sum_{k=1}^\infty \frac{1-e^{ikx}}{1-e^{ix}} \int_0^1 s (k+t)^{-s-1}dt$ $\endgroup$ – reuns Mar 14 at 22:52