Chapter 9.4, Problem 35ES

Use a CAS to confirm that ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{1}{k^2}}=\frac{{\pi }^2}{6}\text{and}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{1}{k^4}=\frac{{\pi }^4}{90}}$

and then use these results in each part to find the sum of the series.

$\left(\text{a}\right){\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{3k^2-1}{k^4}\left(\text{b}\right)}{\displaystyle \underset{k=3}{\overset{\infty }{\sum }}\frac{1}{k^2}\left(\text{c}\right){\displaystyle \underset{k=2}{\overset{\infty }{\sum }}\frac{1}{{\left(k-1\right)}^4}}}$