My main goal is understand useful facts about my computations, that could be wrong, the way shold be too a street without exit looking for a evaluation of Apéry's constant, since I don't use any special method.
In a hand we have that $$\int_0^1\sum_{k=1}^n z^{3k^3-1}dz=\sum_{k=1}^n\int_0^1 z^{3k^3-1}dz=\frac{1}{3}\sum_{k=1}^n\frac{1}{k^3},$$ since the sum is finite, leaving the partial sum of $\frac{1}{3}\zeta(3)$, where $\zeta(3)$ is Apéry's constant, by elementary integration since previous partial sum starts as $$\sum_{k=1}^n z^{3k^3-1}=z^2+z^{23}+z^{80}+z^{191}+z^{374}+\cdots$$
Wolfram Alpha doesn't give me the convergence of corresponding infinite series, neither a closed form for the sum of partial $\sum_{k=1}^n z^{3k^3-1}$. I believe that the convergence holds for infinite series in $|z|<1$.
In the other hand we take the Taylor expansion series around $z=1$ for the monomial $z^{3k^3-1}$ (fixed an integer $k\geq 1$), this
$$z^{3k^3-1}=\sum_{\nu=0}^\infty\binom{3k^3-1}{\nu}(z-1)^{\nu},$$ then if it is possible (I don't know if I have convergence in the following computations, well I've computed with a monomial, and change $\sum^\infty$ with $\int^{z=1}$, where $|z|=1$) $$\int_0^1\sum_{k=1}^n z^{3k^3-1}dz=\sum_{k=1}^{n}\sum_{\nu=0}^\infty\binom{3k^3-1}{\nu}\Big(\frac{(z-1)^{\nu+1}}{\nu +1}\Big|_0^1=-\sum_{k=1}^{n}\sum_{\nu=0}^\infty\binom{3k^3-1}{\nu}\frac{(-1)^{\nu+1}}{\nu +1}.$$
Thus
$$\frac{1}{3}\sum_{k=1}^n\frac{1}{k^3}=-\sum_{\nu=0}^\infty\frac{(-1)^{\nu+1}}{(\nu +1)!}\sum_{k=1}^{n}(3k^3-1)(3k^3-2)\cdots(3k^3-1-\nu+1).$$
Question. a)Can I take the Taylor series for the monomial $z^{3k^3-1}$ around $z=1$ and after change $\sum^\infty$ and $\int^{z=1}$ as my computations does show? b) If there are no mistakes, if feasible give a sum for $\sum_{k=1}^n (3k^3-1)(3k^3-2)\cdots(3k^3-\nu)$? (Since I know that the sum $\zeta(3)$ is very difficult, I look for an alternative identity that I've computed, but my main goal is to know if my computations were justified).
Thanks in advance.