I asked a question that is related to this question and claimed that Generalized Conjecture: $$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(P_r(n) \frac{a \pi}{b}\right) \tag 1 $$
I have a conjecture that if $P_r(n)=\sum\limits_{k = 1 }^ n k^{2r}$ where r is a positive integer,
$f(m)$ function is periodic function when $a,b,m$ positive integers and
$ \sum\limits_{k = 1 }^T f(k)=0 $
where ($T$) is the period value.
(You may see a proof for (Conjecture (1) above) by @FabioLucchini)
I have found out an example and It shows my generalized conjecture can be extended more. I have tested with many numerical values that It supports my extended conjecture below.
$$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(G(n) \frac{a \pi}{b}\right) \tag 2 $$
$$G(n)=\frac{n(n+1)(2n+1)(3n^2+3n-4)}{30}$$ It also satisfies my generalized conjecture $(1)$ above. $G(n)$ can be written as:
$$G(n)=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}-\frac{3}{5}\frac{n(n+1)(2n+1)}{6}=\sum\limits_{k = 1 }^ n k^{4}-\frac{3}{5}\sum\limits_{k = 1 }^ n k^{2}=P_2(n)-\frac{3}{5}P_1(n)$$
Test link for $a=2,b=5$ and $m=500$
The numerical values and my works on the subject estimates the extension of the conjecture above. It is just strong sense without proof that it must be true.
More generalized conjecture can be written:
Extended Conjecture: $$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(\sum\limits_{k = 1 }^\infty \frac{a_k \pi}{b_k}P_k(n) \right) \tag 3 $$
More extended conjecture claims that if $P_r(n)=\sum\limits_{k = 1 }^ n k^{2r}$ where r is a positive integer,
$f(m)$ function is periodic function when $a_k$ is any integers, $b_k$ is non-zero integers and $m$ positive integers.$\sum\limits_{k = 1 }^T f(k)=0 $ where ($T$) is the period value for all possible $a_k,b_k,m$ integers.
Can the extended conjecture be proven? What is the period formula for it?
I also have been still looking for $G(n)$ polynomials that is different from $G(n)=\sum\limits_{k = 1 }^\infty \frac{a_k}{b_k}P_k(n)$ satisfies $ \sum\limits_{k = 1 }^T f(k)=0 $ $\tag{4}$ for all $a,b,m$ positive integers (Please consider Equation $(2)$)
Finally, I claim that there is no any polynomials different from $G(n)=\sum\limits_{k = 1 }^\infty \frac{a_k}{b_k}P_k(n)$ that satisfy $ \sum\limits_{k = 1 }^T f(k)=0 $. Please inform me if you find a counter-example.
Thanks for answers