I am trying to generalize the fact that, for $k>\frac12$,


To reach this I start off with the Fourier series

$$ t^2=\frac{\pi^2}3+4\sum_{n\geq1}\frac{(-1)^n}{n^2}\cos(nt),\qquad |t|\leq\pi\\ \sum_{n\geq1}\frac{(-1)^n}{n^2}\cos(nt)=\frac{t^2}4-\frac{\pi^2}{12} $$

integrate both sides from $0$ to $x$: $$ \sum_{n\geq1}\frac{(-1)^n}{n^2}\int_0^x\cos(nt)dt=\frac{x^3}{12}-\frac{\pi^2x}{12}\\ \sum_{n\geq1}\frac{(-1)^n}{n^3}\sin(nx)=\frac{x^3}{12}-\frac{\pi^2x}{12} $$ plugging in $x=m^{-2k}$ for $m\geq1$, and $k>1/2$, $$\sum_{n\geq1}\frac{(-1)^n}{n^3}\sin(n/m^{2k})=\frac1{12m^{6k}}-\frac{\pi^2}{12m^{2k}}$$ Then applying $\sum\limits_{m\geq1}$ on both sides, $$\sum_{m\geq1}\sum_{n\geq1}\frac{(-1)^n}{n^3}\sin(n/m^{2k})=\frac1{12}\zeta(6k)-\frac{\pi^2}{12}\zeta(2k)$$ With the same process, we have

$$\begin{align} \sum_{m\geq1}\sum_{n\geq1}\frac{(-1)^n}{n^5}\sin(n/m^{2k})&=\frac{\pi^2}{72}\zeta(6k)-\frac1{240}\zeta(10k)-\frac{7\pi^4}{720}\zeta(2k)\\ \end{align}$$ I am trying to find a general form in terms of $\zeta$ values of $$\begin{align} S_j(k)&=\sum_{m\geq1}\sum_{n\geq1}\frac{(-1)^n}{n^j}\sin(n/m^{2k}),\qquad \text{j is odd},\quad j>0\\ \end{align}$$ And as you've seen, I've found up to $j=5$, but I would like to know if a general form exists. Any help is appreciated.

  • $\begingroup$ Sorry, what does $\zeta$ stand for? $\endgroup$ Jan 1, 2019 at 11:08
  • $\begingroup$ @Mike I assume that this should be the Riemann Zeta Function $\zeta(s)$ hence you can see how the $\frac1{4m^{4k}}$ term of the RHS within the line $$\sum_{n\geq1}\frac{(-1)^n}{n^2}\cos(n/m^{2k})=\frac1{4m^{4k}}-\frac{\pi^2}{12}$$ becomes $\frac14\zeta(4k)$ after summing over all integer $m\geq 1$ which equals the defintion of the Riemann Zeta Function for $\operatorname{Re}(4k)>1$. $\endgroup$
    – mrtaurho
    Jan 1, 2019 at 11:56
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    $\begingroup$ @Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times? $\endgroup$
    – mrtaurho
    Jan 1, 2019 at 12:25
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    $\begingroup$ @clathratus: Thanks for that! $\endgroup$ Jan 1, 2019 at 19:48
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    $\begingroup$ How is $$\sum_{m\geq 1}\left( \frac{1}{4m^{2k}} -\frac{\pi^2}{12}\right) $$ convergent? That is what you are summing, right? The first term gives the zeta term but the other term gives $-\infty$ if I'm not mistaken. $\endgroup$
    – Shashi
    Jan 1, 2019 at 22:43

1 Answer 1


Suppose we define the function $$R_{q}(\ell)=\sum_{k\ge1}\sum_{n\ge1}\frac{(-1)^n}{n^{2q+1}}\sin\frac{n}{k^{\ell}}\ .$$

Theorem For $q\in \Bbb N$ and $\ell\in\Bbb N_{>1}$ we have the explicit evaluation $$R_q(\ell)=\frac{(-1)^{q+1}}{2(2q+1)!}\zeta\left((2q+1)\ell\right)+\sum_{i=0}^{q-1}(-1)^i\frac{2^{1-2(q-i)}-1}{(2i+1)!}\zeta(2q-2i)\zeta((2i+1)\ell).\tag{*}$$


To prove this, define the function $$j_q(t)=\sum_{n\ge1}\frac{(-1)^n}{n^{2q+1}}\sin(nt)$$ so that $$R_q(\ell)=\sum_{k\ge1}j_q(k^{-\ell}).\tag{1}$$

We start by noticing that $$\begin{align} \sum_{n\ge1}\frac{(-1)^n}{n^{2q}}(1-\cos(nv))&=\int_0^v j_{q-1}(u)du\\ \sum_{n\ge1}\frac{(-1)^n}{n^{2q}}\cos(nv)&=\mathrm{Li}_{2q}(-1)-\int_0^v j_{q-1}(u)du\\ \sum_{n\ge1}\frac{(-1)^n}{n^{2q}}\cos(nv)&=(2^{1-2q}-1)\zeta(2q)-\int_0^v j_{q-1}(u)du\\ \sum_{n\ge1}\frac{(-1)^n}{n^{2q}}\int_0^t\cos(nv)dv&=(2^{1-2q}-1)\zeta(2q)t-\int_0^t\int_0^v j_{q-1}(u)dudv\\ j_q(t)&=(2^{1-2q}-1)\zeta(2q)t-\int_0^t\int_0^v j_{q-1}(u)dudv. \end{align}$$ And with the well known Fourier series $j_0(t)=-t/2$, it becomes evident from induction that $j_q$ is a polynomial of degree $2q+1$ in $t$ whose exponents are all odd, and $j_q(0)=0$. That being established, we set $$j_q(t)=\sum_{i=0}^{q}\sigma_i^{(q)}r_{2i+1}(t)$$ where $$r_m(t)=\frac{t^m}{m!}\Rightarrow r_m(t)=\int_0^t r_{m-1}(x)dx$$ and $\sigma_i^{(q)}$ are coefficients that are to be evaluated. As we already saw, $\sigma_0^{(0)}=-1/2=2^{-1}-1$. We plug our polynomial formula into our recurrence relation and get that $$\begin{align} j_q(t)&=(2^{1-2q}-1)\zeta(2q)r_1(t)-\sum_{i=0}^{q-1}\sigma_{i}^{(q-1)}\int_0^t\int_0^v r_{2i+1}(u)dudv\\ &=(2^{1-2q}-1)\zeta(2q)r_1(t)-\sum_{i=0}^{q-1}\sigma_{i}^{(q-1)}r_{2i+3}(t). \end{align}$$ Hence the natural definition $\sigma_0^{(q)}=(2^{1-2q}-1)\zeta(2q)$. We preform the index shift $i=0\mapsto i=1$ so that $$j_q(t)=\sigma_0^{(q)}r_1(t)-\sum_{i=1}^{q}\sigma_{i-1}^{(q-1)}r_{2i+1}(t),$$ which gives us the relation $\sigma_i^{(q)}=-\sigma_{i-1}^{(q-1)}$ so that $$j_q(t)=\sum_{i=0}^{q}\sigma_{i}^{(q)}r_{2i+1}(t)$$ is satisfied. Then for $0\le i\le q-1$ we get $$\sigma_i^{(q)}=-\sigma_{i-1}^{(q-1)}=-(-\sigma_{i-2}^{(q-2)})=\dots=(-1)^{i}\sigma_0^{(q-i)}=(-1)^i(2^{1-2(q-i)}-1)\zeta(2q-2i).$$ Then for $q=i$, $$\sigma_i^{(q)}=\sigma_q^{(q)}=(-1)^q\sigma_0^{(0)}=\frac{(-1)^{q+1}}{2}.$$ Thus $$j_q(t)=\frac{(-1)^{q+1}}{2(2q+1)!}t^{2q+1}+\sum_{i=0}^{q-1}(-1)^i\frac{2^{1-2(q-i)}-1}{(2i+1)!}\zeta(2q-2i)t^{2i+1}.$$ Applying $(1)$ and interchanging the order of summation gives $(\,^*\,)$.


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    $\begingroup$ Nice considerations. (+1) --- Only $dt$ above has to be replaced by $dv $. $\endgroup$
    – user90369
    Aug 5, 2019 at 8:25
  • $\begingroup$ Side note: we can also use this approach to show that $$Q_{2n}=\frac{(-1)^n2^{2n-1}}{2^{2n}-1}\left(\frac{1}{2(2n)!}+\sum_{k=1}^{n-1}\frac{(-1)^k(1-2^{1-2k})}{(2n-2k)!}Q_{2k}\right),$$ where $Q_n=\zeta(n)/\pi^n$ $\endgroup$
    – clathratus
    Jun 5, 2020 at 20:17

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