Leonhard Euler was able to calculate the exact sum of the $p$-series with $p=4$: $$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}.$$ Use this fact to find the sum of each series.

$\displaystyle \sum_{n=3}^\infty \frac{1}{n^4} =$

$\displaystyle \sum_{n=3}^\infty \frac{1}{(n+3)^4} =$

$\displaystyle \sum_{n=1}^\infty \frac{1}{(2 n)^4} =$

$\displaystyle \sum_{n=0}^\infty \frac{1}{(2n + 1)^4} =$