Given the sine cardinal function,
$$\rm{sinc}(x) = \frac{\sin x}x$$
for $x\neq0$. We have the nice evaluations,
$$\sum_{k=1}^\infty \rm{sinc}(k) = \sum_{k=1}^\infty \rm{sinc}^2(k)=-\tfrac12+\tfrac12\pi$$
$$\sum_{k=1}^\infty \rm{sinc}^3(k)=-\tfrac12+\tfrac38\pi$$
$$\sum_{k=1}^\infty \rm{sinc}^4(k)=-\tfrac12+\tfrac13\pi$$
$$\sum_{k=1}^\infty \rm{sinc}^5(k)=-\tfrac12+\tfrac{115}{384}\pi$$
$$\sum_{k=1}^\infty \rm{sinc}^6(k)=-\tfrac12+\tfrac{11}{40}\pi$$
then the not-so-nice,
$$\sum_{k=1}^\infty \rm{sinc}^7(k)=-\tfrac12+\quad\\ \tfrac{1}{46080}(129423\pi-201684\pi^2+144060\pi^3-54880\pi^4+11760\pi^5-1344\pi^6+64\pi^7)$$
However, I found this can be prettified as,
$$\sum_{k=1}^\infty \rm{sinc}^7(k)=-\frac12+\frac{7\cdot29^2\,\pi}{2^5\,6!}+\frac{\pi\big(\pi-\tfrac72\big)^6}{6!}$$
Questions:
- Why is the closed-form for $n=7$ far more complicated than $n<7$? (And a good lesson that "patterns" may be illusory.)
- What is $n=8$ in terms of $\pi$? (Maybe also for $n=9$?)
Update: Courtesy of Oliver Oloa's comment, for $n=8$, after some tweaking is,
$$\sum_{k=1}^\infty \rm{sinc}^8(k)=-\frac12+\frac{151\pi}{630}-\frac{\pi\big(\pi-\tfrac82\big)^7}{7!}$$
but $n=9$ is more complicated. See second answer below.
Best Answer
Using Bernoulli polynomials, one can make a general formula: $$S_n=\sum_{k=1}^{\infty}\frac{\sin^n k}{k^n}=-\frac{\pi^n}{2n!}\sum_{k=0}^{n}(-1)^k\binom{n}{k}B_n\left(\Big\{\frac{n-2k}{2\pi}\Big\}\right),$$ where $\{x\}=x-\lfloor x\rfloor$ denotes fractional part of $x$. Say, continuing the examples, $$S_{10}=-\frac{1}{2}-\frac{1093\pi}{672}+\frac{5883\pi^2}{896}-\frac{2449\pi^3}{288}+\frac{563\pi^4}{96}\\-\frac{1423\pi^5}{576}+\frac{43\pi^6}{64}-\frac{103\pi^7}{864}+\frac{3\pi^8}{224}-\frac{\pi^9}{1152}+\frac{\pi^{10}}{40320}.$$ BTW, $n=7$ is the first with $n>2\pi$, which causes the complication.