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I have tried to evaluate this sum $\sum_{n=1}^{\infty}\frac{\sin n \log n}{n}$ using :$\sum_{n=1}^{\infty}\frac{\sin n }{n}$ , Really the partial sum of the titled series given by polylogarithm function and lerch transcendent functions as shown here, But i'm not familiar about convergence of these function for large n , Then my question here is :

What is the exact value of :$$\sum_{n=1}^{\infty}\frac{\sin n \log n}{n}$$ ?

zeraoulia rafik
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  • Honestly I would not expect a closed form to exist. – Szeto Jul 06 '18 at 14:23
  • @gammatester But Wolfy says it is around 0.007. – Szeto Jul 06 '18 at 14:37
  • @szeto: Yes, it seems to be non-zero, and therefore I deleted my comment. I have plotted with Maple because here WA does give a result. Plotting up to 2000 shows that it is slightly above 0. – gammatester Jul 06 '18 at 14:44

1 Answers1

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$0<x<1$ : $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{\ln n}{n}\sin(2\pi n x) = \frac{\pi}{2}\left(\ln\frac{\Gamma(x)}{\Gamma(1-x)}-(1-2x)(\gamma + \ln(2\pi))\right)$

see e.g. E. E. Kummer, page 1 to 4

See also here , look for the first derivation for $t$ of $B_t(x)$ , a generalization of the Bernoulli polynomial and it’s Fourier series.

It follows:

$$\sum\limits_{n=1}^\infty \frac{\sin n\ln n}{n} = \frac{\pi}{2}\left(\ln\frac{\Gamma\left(\frac{1}{2\pi}\right)}{\Gamma\left(1-\frac{1}{2\pi}\right)}-\left(1-\frac{1}{\pi}\right)(\gamma + \ln(2\pi))\right)$$

user90369
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