<s>[INST] How do I use Euler's result to find the sum of a series? So I am given:
$$ \zeta(4) = \sum_{n=1}^\infty {1\over n^4}={\pi^4 \over 90} $$
I need to use it to find the sum of the following series using the above information.
$$ \sum_{k=1}^\infty {1\over{(k+2)^4}} $$
So, this is what I have so far:
$$ \sum_{k=1}^\infty {1\over{(k+2)^4}} = \sum_{k=3}^\infty {1\over{k^4}}$$
but that is all I have... How do I get rid of the $k=3$? 
 [/INST] Compare $$\sum_{k=3}^{\infty}\frac{1}{k^4}=\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\cdots$$
with$$\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{1}{1^4}+\frac{1}{2^4}+\color{red}{\frac{1}{3^4}+\frac{1}{4^4}+\cdots}.$$
Hence we have $$\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{1}{1^4}+\frac{1}{2^4}+\sum_{k=3}^{\infty}\frac{1}{k^4}\iff \sum_{k=3}^{\infty}\frac{1}{k^4}=\left(\sum_{k=1}^{\infty}\frac{1}{k^4}\right)-\frac{1}{1^4}-\frac{1}{2^4}.$$
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