It is well known that $$ \sum _1 ^\infty \frac {1}{n^2} = \frac {\pi ^2}{6}$$ and $$ \sum _1 ^\infty \frac {1}{n^4} = \frac {\pi ^4}{90}$$
We also know that $$ \sum _1 ^\infty \frac {1}{n^3} $$
$$=1.202056903159594285399738161511449990764986292340498881.....$$
My question is:
Do we have a closed form for this series besides $$ \sum _1 ^\infty \frac {1}{n^3} = \zeta (3) ?$$
Apéry's Constant
As mentioned in comments $$ \sum_{n=1}^\infty\frac1{n^3}=\zeta(3)\tag1 $$ is also known as Apéry's Constant. There is no closed form in terms of a rational multiple of an integer power of $\pi$.
We also have $$ \sum_{n=1}^\infty\frac1{(2n-1)^3}=\frac78\zeta(3)\tag2 $$
However, as shown in this answer, if we alternate the series in $(2)$, we get $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}\tag3 $$ which is also known as $\beta(3)$, the Dirichlet beta function.
Computation of $\boldsymbol{\zeta(3)}$
The series in $(1)$ converges very slowly. To get the sum to about $10$ places, we would need to sum $100000$ terms.
If we use the Euler-Maclaurin Sum Formula, we get that $$ \sum_{k=1}^n\frac1{k^3}=\small\zeta(3)-\frac1{2n^2}+\frac1{2n^3}-\frac1{4n^4}+\frac1{12n^6}-\frac1{12n^8}+\frac3{20n^{10}}-\frac5{12n^{12}}+O\!\left(\frac1{n^{14}}\right)\tag4 $$ Using $n=10$ in $(4)$, we get $$ \sum_{k=1}^\infty\frac1{k^3}=1.2020569031596\tag5 $$