# Generalizing $\sum\limits_{m\geq1}\sum\limits_{n\geq1}\frac{(-1)^n}{n^3}\sin(n/m^{2k})=\frac1{12}\zeta(6k)-\frac{\pi^2}{12}\zeta(2k)$

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

$$\sum_{m\geq1}\sum_{n\geq1}\frac{(-1)^n}{n^3}\sin(n/m^{2k})=\frac1{12}\zeta(6k)-\frac{\pi^2}{12}\zeta(2k)$$

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.

• Sorry, what does $\zeta$ stand for? Jan 1, 2019 at 11:08
• @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 $\operatorname{Re}(4k)>1$. Jan 1, 2019 at 11:56
• @Did Why did you only changed it in the first line at not beneath where the same notation is used seven more times? Jan 1, 2019 at 12:25
• @clathratus: Thanks for that! Jan 1, 2019 at 19:48
• 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. Jan 1, 2019 at 22:43

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 $$R_q(\ell)=\frac{(-1)^{q+1}}{2(2q+1)!}\zeta\left((2q+1)\ell\right)+\sum_{i=0}^{q-1}(-1)^i\frac{2^{1-2(q-i)}-1}{(2i+1)!}\zeta(2q-2i)\zeta((2i+1)\ell).\tag{*}$$

Proof

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 $$(\,^*\,)$$.

$$\square$$

• Nice considerations. (+1) --- Only $dt$ above has to be replaced by $dv$. Aug 5, 2019 at 8:25
• 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 $Q_n=\zeta(n)/\pi^n$ Jun 5, 2020 at 20:17