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Can we do the sums

\begin{align} \sum_{n=1}^{\infty} \frac{(\pm1)^n}{n^{5/2}} \sin(nx)\\ \sum_{n=1}^{\infty} \frac{(\pm1)^n}{n^{5/2}} \cos(nx) \end{align}

They appear similar to known results e.g. this answer, this one and this one, but I can't solve it myself or find an answer.

From a numerical investigation, I find e.g., $$ \sum_{n=1}^{\infty} \frac{1}{n^{5/2}} \cos(nx) \approx - 3/4 \pi |\sin(x/2)| + \pi/2 $$

$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{5/2}} \cos(nx) \approx - 3/4 \pi |\cos(x/2)| + \pi/2 $$ $$ \sum_{n=1}^{\infty} \frac{(\pm1)^n}{n^{5/2}} \sin(nx) \approx \pm \sin(x) $$ Are these results exact? If not, what are the answers? and why are these approximations rather good?

innisfree
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  • Most of such known cases reduce to the Fourier series of the (periodic) Bernoulli polynomials. For these case, we are out of luck and I suspect they may not have a closed form. – Sangchul Lee Jul 05 '17 at 08:39
  • I'm checking my guesses to high order in Mathematica... they don't appear to be exact. – innisfree Jul 05 '17 at 08:46

1 Answers1

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Only a hint:

Using the considerations in The sum of fractional powers $\sum\limits_{k=1}^x k^t$. I only found

$$\sum\limits_{k=1}^\infty\frac{1}{k^{5/2}}(\sin(xk)+\cos(xk))=-\frac{(2\pi)^2}{3}\zeta(-\frac{3}{2},\frac{x}{2\pi})$$

and therefore

$$\sum\limits_{k=1}^\infty\frac{1}{k^{5/2}}\cos(xk)=-\frac{(2\pi)^2}{6}\left(\zeta(-\frac{3}{2},\frac{x}{2\pi})+\zeta(-\frac{3}{2},-\frac{x}{2\pi})\right)$$

and

$$\sum\limits_{k=1}^\infty\frac{1}{k^{5/2}}\sin(xk)=-\frac{(2\pi)^2}{6}\left(\zeta(-\frac{3}{2},\frac{x}{2\pi})-\zeta(-\frac{3}{2},-\frac{x}{2\pi})\right)$$

so that one can compare $\enspace\displaystyle -\frac{(2\pi)^2}{6}\zeta(-\frac{3}{2},\frac{\pm x}{2\pi})\enspace$ with the assumptions for

$\displaystyle \sum\limits_{k=1}^\infty\frac{\sin(xk)}{k^{5/2}}\enspace$ and $\enspace\displaystyle \sum\limits_{k=1}^\infty\frac{\cos(xk)}{k^{5/2}}$ .

To calculate the Hurwitz Zetafunction $\,\zeta(s,x)\,$ especially for negative $\,s\,$ see

the different possibilities here: $\enspace$ https://en.wikipedia.org/wiki/Hurwitz_zeta_function

user90369
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  • Thanks, isn't this a solution for 2 of the sums since I can identify even/odd parts of zeta with cos/sin sums – innisfree Jul 05 '17 at 09:47
  • @innisfree : If $,\displaystyle -\frac{(2\pi)^2}{3}\zeta(-\frac{3}{2},\frac{x}{2\pi})\neq \sin(x)-\frac{3\pi}{4}|\sin\frac{x}{2}|+\frac{\pi}{2},$ (a test with one valid value $x$ is enough) then at least you know that your assumptions can not be an exact result. But it doesn’t mean here, that all are wrong, only that at least one is wrong. – user90369 Jul 05 '17 at 10:22
  • Yes. But I can test each one individually by identifying even/odd parts. – innisfree Jul 05 '17 at 12:02
  • I know what you mean but first I was not sure if is possible to substitute $x$ by $-x$ . But now: it works because of $|x|<2\pi$ . – user90369 Jul 05 '17 at 12:25
  • The wiki link is helpful. Your answer looks like a s=5/2 case of Hurwitz's formula – innisfree Jul 05 '17 at 23:31
  • Not only "looks like", indeed it is. :-) This makes it possible to discuss such series by using the characteristics of the Hurwitz zeta function. --- You can also check characteristics of polylogarithm https://en.wikipedia.org/wiki/Polylogarithm . – user90369 Jul 06 '17 at 08:15