I think that will not be useful to compute the Apéry's constant as $$\zeta(3)=\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{n^2}\int_0^{\frac{\pi}{2n}}\sin^2(3x)dx+\frac{1}{3\pi}\left(\sum_{n=1}^{\infty}\frac{\sin(\frac{3\pi}{n})}{n^2}\right),$$ which is easily deduced from $$\int_0^{\frac{\pi}{2n}}\sin^2(3x)dx=\frac{1}{12}\left(\frac{3\pi}{n}-\sin(\frac{3\pi}{n})\right).$$

For deduce this last identity I've used an online tool of symbolic calculus. I say that couldn't be useful since neither I don't know how evaluate an alternative of the series of integrals $\sum_{n\geq 1} \frac{1}{n^2}\int_0^{\frac{\pi}{2n}}\sin^2(3x)dx$ (if it is feasible, to compute in a distinct form).

In any case I would like to know

Question. It is know, or it is possible to get a closed-form for $$\sum_{n=1}^{\infty}\frac{\sin(\frac{3\pi}{n})}{n^2}$$ in terms of known constants? Thanks in advance.

When I've do more computations, using some trigonometric tricks I've found also with $\sum_{n=1}^{\infty}\frac{\sin(\frac{\pi}{n})}{n^2}$ and $\sum_{n=1}^{\infty}\frac{\sin(\frac{2\pi}{n})}{n^2}$. It is easy to check the absolute convergence of such series, thus by comparision test these series are convergents. Using another time Wolfram Alpha (its Online Series Calculator) I don't to get a closed-form (only an approximation is given for such series) in terms of known constant, this is as exact value. Are know these series? When I've used this computation in its (General) Calculator, yes then I've obtained some closed-form, but the Series Calculator doesn't sure any closed-form.

My attemps were using partial summation and Euler-MacLaurin (first approximation), but I believe that I can not find this real value with these methods.

  • $\begingroup$ I recoignize that the identity involving previous constant is vague, but now I am interesting in if it is possible obtain a closed form for previous cited series and how obtain it. Thank.s $\endgroup$
    – user243301
    Commented Dec 30, 2015 at 16:54
  • 5
    $\begingroup$ Since $e^{ix}=\cos x+i\sin x,$ this is equivalent to asking for the general closed form of $\sum\limits_{n=1}^\infty\dfrac{\sqrt[n]a}{n^2}.$ $\endgroup$
    – Lucian
    Commented Dec 30, 2015 at 18:36
  • 1
    $\begingroup$ Maybe it would be useful to consider: $\sum\limits_{n=1}^{\infty}\frac{sin^2(\alpha\cdot n)}{n^2\cdot\alpha}=\frac{\pi-\alpha}{2}$ $\endgroup$
    – user515010
    Commented Jan 7, 2018 at 15:20
  • 1
    $\begingroup$ I get $-0.046998999668291721739\dots$ for the series in question, which is very close to a rational number $-0.047$ though it's probably just a coincidence $\endgroup$
    – Yuriy S
    Commented Apr 11, 2018 at 9:09
  • 1
    $\begingroup$ Could you clarify this part: "When I've used this computation in its (General) Calculator, yes then I've obtained some closed-form" - what kind of closed form did you obtain and how? $\endgroup$
    – Yuriy S
    Commented Apr 11, 2018 at 11:35

1 Answer 1


I wouldn't expect a closed form, however we can find an integral form for the series involving Kelvin functions:

$$f(a)=\sum_{n=1}^\infty \frac{\sin \left( \frac{a}{n} \right)}{n^2}=-\frac{1}{2a^2} \int_0^\infty \frac{\text{ber}_1 ( \sqrt{2}~u)+\text{bei}_1 (\sqrt{2}~u) }{e^{\frac{1}{2a}u^2}-1} u^2 du \tag{1}$$

Where the Kelvin functions can be defined as real and imaginary parts of a certain Bessel function:

$$\text{ber}_1 (\sqrt{2}~u)=\Re J_1 \left((i-1)u \right)=\frac{1}{2\pi} \int_{-\pi}^\pi e^{u \sin t} \cos \left(t+u \sin t \right) dt$$

$$\text{bei}_1 (\sqrt{2}~u)=\Im J_1 \left((i-1)u \right)=\frac{1}{2\pi} \int_{-\pi}^\pi e^{u \sin t} \sin \left(t+u \sin t \right) dt$$

The form is derived by directly using the Taylor series for $\sin$ and reversing order of summation:

$$f(a)=\sum_{n=1}^\infty \sum_{k=0}^\infty \frac{(-1)^k a^{2k+1}}{(2k+1)! n^{2k+3}}=\sum_{k=0}^\infty \frac{(-1)^k a^{2k+1}}{(2k+1)!} \zeta (2k+3)$$

Now we use the integral form of the zeta function:

$$\zeta (2k+3)=\frac{1}{(2k+2)!} \int_0^\infty \frac{x^{2k+2}}{e^x-1}dx$$


$$f(a)=\frac{a}{2} \int_0^\infty \frac{x^2dx}{e^x-1} \sum_{k=0}^\infty \frac{(-1)^k (ax)^{2k}}{(2k+1)!^2 (k+1)}$$

The sum in terms of Kelvin functions was found by Wolfram Alpha. Then after some changes of variable in the integral, we get the form (1).

  • $\begingroup$ Many thanks I accept the answer and reasoning, since I was asking if is it possible to get a closed-form. On the other hand your formulas have a great mathematical beauty, many thanks I am going to study it in detail. $\endgroup$
    – user243301
    Commented Apr 11, 2018 at 15:07

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