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 $k=3$?
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 $$\sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{1}{1^4}+\frac{1}{2^4}+\sum_{k=3}^{\infty}\frac{1}{k^4}\iff \sum_{k=3}^{\infty}\frac{1}{k^4}=\left(\sum_{k=1}^{\infty}\frac{1}{k^4}\right)-\frac{1}{1^4}-\frac{1}{2^4}.$$
