Can this approximate closed form of Apery's constant $\zeta(3)$ be improved? I know that an approximate closed form is not really a solution. However, I would like to present a method that gives a closed form of $\zeta(3)$ that is accurate to the 5th decimal, hoping that it may help to find the exact expression 
Consider a real function:
$g(x)=-\frac{1}{2!}+\frac{x^3}{5!}-\frac{x^6}{8!}+\frac{x^{9}}{11!}-...$
This function $g(x)$ has infinite positive roots. Let's denote these roots by $r_i$. Because $g(x)$ is a function of $x^3$ every root $r_i$ should be a triple root. We can show that:
$g(x)=-\frac{1}{2}\prod_{i=1}^{\infty}{(1-\frac{x^3}{r_i^3})}$
On the other hand, roots of $g(x)=0$ are given by:
$2\sin(\frac{\pi}{6}-\frac{\sqrt{3}}{2}x)=e^{-\frac{3x}{2}}$. As $x$ increases, the exponential part assimptotes to zero and roots of the function become closer to the roots of a sine wave $2\sin(\frac{\pi}{6}-\frac{\sqrt{3}}{2}x)=0$. The roots of sine equation are easy to show:$\mu_i=\frac{(6i+1)\pi}{3\sqrt{3}}, i=1,2,3,...  $
Now, assuming that these two sets of roots ($r_i$ and $\mu_i$)  are close to eachother, we can write:
$g(x)\cong-\frac{1}{2}\prod_{i=1}^{\infty}(1-\frac{x^3}{\mu_i^3})$
 or 
$g(x)\cong-\frac{1}{2}\prod_{i=1}^{\infty}(1-\frac{(3\sqrt{3})^3x^3}{(6i+1)^3\pi^3})$
The coefficient for the $x^3$ must be equal to that of the infinite series of $g(x)$. 
$\sum_{i=1}^{\infty}\frac{(3\sqrt{3})^3x^3}{(6i+1)^3\pi^3}\cong\frac{2x^3}{5!}$
and $\sum_{i=1}^{\infty}\frac{1}{(6i+1)^3}\cong\frac{2\pi^3}{5!(3\sqrt{3})^3}$ and we know that $\sum_{i=1}^{\infty}\frac{1}{(6i+1)^3}=\frac{(91\zeta(3)+2\sqrt{3}\pi^3)}{216}-1$ combining these two equations gives us an approximate closed form of $\zeta(3)$:
$\zeta(3)\cong\frac{4(7290-67\sqrt{3}\pi^3)}{12285}$
which is accurate to the fifth decimal. This method is analogous to Euler's derivation of $\zeta(2)$ except that the roots of this function $g(x)$ are slightly nonlinear. Is it possible to reduce the error in the approximation of the roots $r_i$ by $\mu_i$?
 A: First of all, thank you for the highly interesting question, it provides an excellent exercise!

Unfortunately it is not possible to improve the approximation by your method alone. Of course, the error in your approximation is due to taking the roots of $2\sin(\frac{\pi}{6}-\frac{\sqrt{3}}{2}x)=e^{-\frac{3x}{2}}$ when $x\to\infty$ and thus using the asymptotic $2\sin(\frac{\pi}{6}-\frac{\sqrt{3}}{2}x)=0$ instead. In order to alter this error, we examine your function $g(x)$:
$$
g(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{3n}}{(3n+2)!}=\frac{e^{-x}\left(2e^{3x/2}\sin(\frac{1}{6}(\pi-3\sqrt3 x))-1\right)}{3x^2},
$$
and note that in order to change the properties of the sine wave on the RHS we need to alter the factorial on the LHS. Let $k$ now denote a positive integer $\geq0$ and examine what happens when we change $(3n+2)!$ to some $(3n+k)!$. In particular, we examine the three cases $k\equiv0 \ (\textrm{mod} \ 3)$, $k\equiv1 \ (\textrm{mod} \ 3)$ and $k\equiv2 \ (\textrm{mod} \ 3)$.

Case 1. $k=3j$, $j\in\mathbb{Z}$.
Let's examine $k=0$ first and generalize later.
$$
g'(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{3n}}{(3n)!}=-\frac13 e^{-x} \left(2e^{3x/2}\cos\left(\frac{\sqrt{3} x}{2}\right)+1\right),
$$
and by similar methods we get $\zeta(3)\cong \frac87\left(\frac{\pi^{3}}{6(3)^{3/2}}\right)=1.136602043\ldots$ which is not a very good approximation. Having $j>0$ yields similar results that include the cosine function, but with some leading $\exp$ terms. For instance, we have for $j=3$:
$$
\sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{3n}}{(3n+9)!}=\frac{e^{-x}\left(-e^{x}x^6+120e^xx^3-720e^x+480e^{3x/2}\cos(\frac{\sqrt{3} x}{2})+240\right)}{720x^9}
,$$
and we can see that when $x\to\infty$ we are making an even bigger error in our approximation by assuming that its roots are those that satisfy the equation
$$
480\cos\left(\frac{\sqrt{3} x}{2}\right)=0.
$$
In fact, as $j$ increases so does our error.

Case 2. $k=3j+1$, $j\in\mathbb{Z}$.
In this case we get similar expressions as your original sine wave, but with some differences in sign (pun not intended). For $j=0$ we have
$$
g'(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{3n}}{(3n+1)!}=-\frac{e^{-x}\left(2e^{3x/2}\sin(\frac{1}{6}(\pi+3\sqrt3 x))-1\right)}{3x^2},
$$
which has the same roots as your original function $g(x)$ and the same error in approximating $\zeta(3)$. As in Case 1, increasing the value of $j$ leads to even more leading $\exp$ terms and thus increases the error in our approximation.

Case 3. $k=3j+2$, $j\in\mathbb{Z}$.
The case $j=0$ corresponds to your original function $g(x)$. As with cases 1 and 2, increasing $j$ increases the amount of leading $\exp$ terms and thus the error in the approximation due to taking the roots of the asymptotic value when $x\to\infty$.

It remains to be seen whether it is possible to sensibly generalize the factorial to $(in+j)!$, $i,j\in\mathbb{Z}, i>0$ in order to get other approximations for other odd zeta values, perhaps several of these cases might yield good approximations. Euler himself tried to generalize his proof of the Basel problem to compute values for $\sum_n n^{-(2k+1)}$, $k>0$ but found no way to achieve this. Since then, considerable brain power has been expended to do just that, without success. I believe that is a good indicator that this path might not lead to a nice closed-form expression for $\zeta(3)$ in terms of $\pi^3$.

EDIT 11.12.2018
Taking the function
$$
\sum_{n=0}^{\infty} \frac{(-1)^{n+1}x^{5n}}{(3n+2)!}=\frac{e^{-x^{5/3}}\left(2e^{3x^{5/3}/2}\sin(\frac{1}{6}(\pi-3\sqrt3 x^{5/3}))-1\right)}{3x^{10/3}}
$$
yields the same approximation as $g(x)$, i.e., with $x^{3n}$ instead of $x^{5n}$ in the sum. This provides further indication that this approach is unlinkely to yield any better results on $\zeta(3)$.
