For $n\geq 1$, $\sum_{k=0}^{\infty}\frac{(-1)^{k}(nk+1)^{3}}{(k+1)^6}$ in terms of $\zeta(3)$ and $\zeta(5)$ from a series calculator. Is possible?

Question

I am doing experiments with this widget (Wolfram Alpha, a Series calculator, by HIghOPS)

http://www.wolframalpha.com/widgets/view.jsp?id=86ceba9f35c96ebae137e44a36c7261a

and take for

Example. That $$\sum_{n=0}^{\infty}\frac{(-1)^n (5n+1)^3}{(n+1)^6}=\frac{354375\zeta(3)+850500\zeta(5)-11025\pi^4-248\pi^6}{3780}$$

when I type

Sequence (n is the variable) $\qquad\qquad$ (-1)^n(5n+1)^3/(n+1)^6

Start value $\quad\qquad\qquad\qquad\qquad\qquad$ 0

End value $\quad\qquad\qquad\qquad\qquad\qquad$ inf

and I am susprising because the first examples for integers $n\geq 1$ of

$$\sum_{k=0}^{\infty}\frac{(-1)^{k}(nk+1)^{3}}{(k+1)^6}$$ are given in a closed form only in terms of $\zeta(3)$ and $\zeta(5)$ (I say withot involve distinct particular values of Riemann zeta function), and the ability to compute this series values.

Question. Is known or is easy deduce a formula to compute $\sum_{k=0}^{\infty}\frac{(-1)^{k}(nk+1)^{3}}{(k+1)^6}$ only in terms of $\zeta(3)$ and $\zeta(5)$, and known reals (as powers of $\pi$), when $n\geq 1$ is a fixed integer? Am I wrong in the use of this calculator? Thanks in advance.

Answer 1

Write $(5k+1)^3 = a(k+1)^3+b(k+1)^2+c(k+1)+d$. Then simplify, and you get:

$$\sum_{k} (-1)^k\left(\frac{a}{(k+1)^3} + \frac{b}{(k+1)^4}+\frac{c}{(k+1)^5}+ \frac{d}{(k+1)^6}\right)$$

Now you just need to write $\sum\frac{(-1)^{k-1}}{k^s}$ in terms of $\zeta(s)$.

But it is relatively easily shown that this is $\zeta(s)\left(1-\frac{1}{2^{s-1}}\right)$.

So you end up with:

$$\frac{3a}{4}\zeta(3) + \frac{7b}{8}\zeta(4) + \frac{15c}{16}\zeta(5)+\frac{31d}{32}\zeta(6)$$

If we write $5k+1=5(k+1)-4$, we see that $$a=5^3,b=3\cdot 5^2\cdot 4, c=3\cdot 5\cdot 4^2, d=4^3$$

A little tinkering shows that $\frac{3}{4}\cdot 5^3$ is, indeed, equal to $$\frac{354375}{3780}$$

This means that the sum is actually in terms of $\zeta(3),\zeta(4),\zeta(5),$ and $\zeta(6)$. But we know exact values for $\zeta(4)$ and $\zeta(6)$, so Wolfram is just making that substitution. Exact values are not known for the odd $\zeta(n)$, so Wolfram just leaves them in as $\zeta(n)$.