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The series $\sum_{n=1}^\infty \frac{\sin n }{\sqrt{n}}$ is clearly convergence as can be shown with the Dirichlet's test. But what is this value and how to evaluate this sum to get a closed form solution?

With WolframAlpha I arrive for example at:

$$\dfrac{1}{2} i (Li_{1/2}(e^{-i}) - Li_{1/2}(e^i))≈1.04398 $$

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  • $\begingroup$ Most series don't have a closed-form, this is one of them $\endgroup$
    – reuns
    Commented Dec 14, 2019 at 4:55

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This is a Clausen function of order 1/2 given by

$$\operatorname{Si}_{1/2}(1)=\sum_{n=1}^\infty\frac{\sin(n)}{\sqrt n}$$

By applying Euler's formula for $\sin(z)=\frac1{2i}(e^{iz}-e^{-iz})$ one gets the relation to the polylogarithm:

$$\operatorname{Si}_{1/2}(1)=\frac1{2i}(\operatorname{Li}_{1/2}(e^i)-\operatorname{Li}_{1/2}(e{-i}i))$$

Quicker numerical evaluation can be done using various relations such as the above. Applying a more brute force approach, one may take the Cesaro summation:

$$\operatorname{Si}_{1/2}(1)=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sum_{j=1}^k\frac{\sin(j)}{\sqrt j}$$

Evaluating the above at $n=33208$ gives us

$$\operatorname{Si}_{1/2}(1)\simeq1.0439773797177627$$

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  • $\begingroup$ Being "surprised" by your result, I made a few calculations. I think that I never saw something slower. Cheers :-) $\endgroup$ Commented Dec 14, 2019 at 15:55
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Another possible form is $$\sum_{n=1}^\infty \frac{\sin (n)}{\sqrt{n}}=\Im\left(\sum_{n=1}^\infty \frac{e^{in}}{\sqrt{n}}\right)=\Im\left(\text{Li}_{\frac{1}{2}}\left(e^{-i}\right)\right)$$ which can rewrite as $$\sum_{n=1}^\infty \frac{\sin (n)}{\sqrt{n}}=\Im\left(\frac{(-1)^{1/4} \zeta \left(\frac{1}{2},1-\frac{1}{2 \pi }\right)+(-1)^{3/4} \zeta \left(\frac{1}{2},\frac{1}{2 \pi }\right)}{i\sqrt{2} }\right)$$ and its numerical value is $$1.04398210284916152745329487258675045979090714472261220374895285877066908586$$ which is quite close to $$\frac{1+2\log (5)}{20 (\zeta (3)-1)}=1.043982105$$

Edit

What looked interesting (at least to me) was to compute the partial sum $$S_k=\sum_{n=1}^{10^k}\frac{\sin (n)}{\sqrt{n}}$$ Using double precision, some results $$\left( \begin{array}{cc} 1 & 1.2080477203 \\ 2 & 0.9400223327 \\ 3 & 1.0407653236 \\ 4 & 1.0511687406 \\ 5 & 1.0469310313 \\ 6 & 1.0429497497 \\ 7 & 1.0443111845 \\ 8 & 1.0440619434 \\ 9 & 1.0439664828 \\ \cdots & \cdots \\ \infty & 1.0439821028 \end{array} \right)$$

Quite slow, isn't it ?

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  • $\begingroup$ You don't want to compute the series at powers of tens, but rather at near multiples of $2\pi$. It converges a tad faster if you choose your bound like so. $\endgroup$ Commented Dec 14, 2019 at 17:05
  • $\begingroup$ @SimplyBeautifulArt. I made it that way on purpose. Cheers :-) $\endgroup$ Commented Dec 15, 2019 at 8:42

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