I have the sum $$ \sum_{n=1}^\infty \dfrac{\cos (n \theta)}{n^5} = \dfrac{\text{Li}_5 (e^{i\theta}) + \text{Li}_5 (e^{-i\theta})}{2}, $$ where $0\leq\theta < 2 \pi$ is an angle and $\text{Li}_5(z)$ is a polylogarithm function. I want to know the dominant behaviour of my function. I mostly care for the regime $\theta \ll 1$. Could someone please help me find some information on this? I've mostly been only able to find behavours for $z \approx 0$ and for $|z|<1$, which are not my case!
$\begingroup$
$\endgroup$
4
-
$\begingroup$ For real $\theta$ your function seems to evaluate fairly close to $\cos(\theta)$ (see WolframAlpha evaluation). $\endgroup$– Steven ClarkCommented Mar 25 at 0:06
-
1$\begingroup$ $\theta=0$ corresponds to $z=1$ which is where the polylogs conventionally have a branch point, which may be the source of your difficulty. On the other hand, if $S(\theta)=\sum_nn^{-5}\cos(n\theta)$, then we can see directly that $S \sim \zeta(5)-\theta^2\frac{\zeta(3)}{2}$ as $\theta\to 0$, where $\zeta$ is the Riemann zeta function. The quartic term cannot be obtained as easily. These first two terms are very close to the expansion of $\cos$ because $\zeta(5)$ and $\zeta(3)$ are numerically 'close' to unity $\endgroup$– SalCommented Mar 25 at 1:32
-
$\begingroup$ @GabrielYbarraMarcaida Did I answer your question? $\endgroup$– GaryCommented Mar 26 at 23:47
-
1$\begingroup$ Yes, sorry, forgot to tick it. Many thanks!!! $\endgroup$– Gabriel Ybarra MarcaidaCommented Mar 27 at 0:00
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
It is known that $$ \operatorname{Li}_5 ({\rm e}^\mu ) = - \frac{1}{24}\mu ^4\log ( - \mu ) + \frac{{25}}{{288}}\mu ^4 + \sum\limits_{k = 0,k \ne 4}^\infty {\frac{{\zeta (5 - k)}}{{k!}}\mu ^k } $$ for $\left| \mu \right| < 2\pi$, $\mu\neq 0$. Thus, \begin{align*} \sum\limits_{n = 1}^\infty {\frac{{\cos (n\theta )}}{{n^5 }}} = \operatorname{Re}(\operatorname{Li}_5 ({\rm e}^{ - {\rm i}\theta } )) = - \frac{1}{24}\theta ^4\log \theta & + \zeta (5) - \frac{{\zeta (3)}}{2}\theta ^2 + \frac{{25}}{{288}}\theta ^4 \\ & + \sum\limits_{k = 3}^\infty {( - 1)^k \frac{{\zeta (5 - 2k)}}{{(2k)!}}\theta ^{2k} } \end{align*} for $0<\theta<2\pi$.
