The solution to $\sum_{n=1}^\infty\frac1{n^3}$ in a form not considered closed form? Is there a solution to $\sum_{n=1}^\infty\frac1{n^3}$ that isn't in the standard closed form?  I was wondering if a looser definition of "solution" could allow someone to solve this or if the solution already exists.
Obviously $\zeta(3)$ doesn't cut it, it must be a solution that I can use to calculate the series without using a series directly equal to the one I'm trying to solve.
Or close bounds that allow squeeze theorem to be applied, as some methods use to solve $\zeta(2)$?  Close bounds findable through calculus based methods?
 A: I don't know if this helps, but there are other representations of the Riemann-Zeta Function that permit faster convergence rates than the representation afforded by the series
$$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$$
One such representation is provided in terms of the alternating series representation of the Dirichlet Eta 
$$\begin{align}
\zeta(s)&=\bbox[5px,border:2px solid #C0A000]{\left(\frac{1}{1-2^{1-s}}\right)\eta(s)}\\\\
&=\left(\frac{1}{1-2^{1-s}}\right)\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \tag 1
\end{align}$$
For $s=3$, we see from $(1)$ that $\zeta(3)$ can be expressed as
$$\bbox[5px,border:2px solid #C0A000]{\zeta(3)=\frac43 \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}} \tag 2$$
Note that $(3)$ provides a rapidly converging series with a built-in error bound on the partial sums provided by the absolute value of the next term of the series.  Thus, we can immediately find the number of terms in the series required to yield an approximation within a specified error.
For example, if we want an approximation with error less than $10^{-5}$, then we can choose the number of terms $N-1$ such that $N>\left(\frac43 10^5\right)^{1/3}\approx 51.087$.  Therefore, by summing the first $51$ terms of the series in $(2)$ we are guaranteed that the result will have an error less than $10^{-5}$.  
Similarly, if we want an approximation with an error less than $10^{-6}$, then we can choose the number of terms $N-1$ such that $N> \left(\frac43 10^5\right)^{1/3}\approx 110.064$.  Therefore, by summing the first $110$ terms of the series in $(2)$ we are guaranteed that the result will have an error less than $10^{-6}$.  

Other alternating series representations given here provide even faster convergence rates than $(2)$.

