Working in the series above and using maclaurin euler formula i get $$\sum _{n=1}^{\infty } \frac{\sqrt{2 \pi } \sqrt{\frac{2 \pi n}{x \sqrt{\frac{4 \pi ^2 n^2}{x^2}-1}}+1}}{\sqrt[4]{x^2-4 \pi ^2 n^2}}+\frac{\sqrt{\frac{\pi }{2}}}{\sqrt{x}}=\sum _{n=1}^{\infty } \frac{1}{\sqrt{k}} \cos (k x)$$ but series diverge , it maybe something wrong?
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1There is no $n$ term inside the summation sign? – BAYMAX Jul 16 '17 at 09:08
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Are you sure about $\sqrt[4]{x^2-4 \pi ^2 n^2}$ in the denominator ? It becomes purely imaginary for real $x$ when $|x|<2\pi n$ . – user90369 Jul 16 '17 at 10:20
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The EML formula cannot be applied to everything. It does not work for $\sum_{n\geq 1}e^{-n^2}$, for instance. – Jack D'Aurizio Jul 16 '17 at 16:30
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I am not sure that your approach will get you the result.
This is my hint. The given series does not converge if $x$ is a integer multiple of $2\pi$, i.e $x=2\pi m$ with $m\in\mathbb{Z}$ because $$\sum _{k=1}^{\infty } \frac{\cos (k 2\pi m)}{\sqrt{k}} =\sum _{k=1}^{\infty } \frac{1}{\sqrt{k}}=+\infty$$ Otherwise, note that (see Prove $\frac{1}{2} + \cos(x) + \cos(2x) + \dots+ \cos(nx) = \frac{\sin(n+\frac{1}{2})x}{2\sin(\frac{1}{2}x)}$ for $x \neq 0, \pm 2\pi, \pm 4\pi,\dots$) $$\sum_{k=1}^n\cos(kx)= \frac{\sin((2n+1)x/2)}{2\sin(x/2)}-\frac{1}{2}$$ and use Dirichlet's Test.
Robert Z
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