Answer
(a) $\frac{9\pi^{4}}{10}$
(b) $\frac{\pi^{4}}{90}-\frac{17}{16}$
Work Step by Step
(a) $\Sigma_{n=1}^{\infty}(\frac{3}{n})^{4}=\Sigma_{n=1}^{\infty}\frac{3^{4}}{n^{4}}=\Sigma_{n=1}^{\infty}\frac{81}{n^{4}}$
$=81\Sigma_{n=1}^{\infty}\frac{1}{n^{4}}$
$=81\times\frac{\pi^{4}}{90}$
$=\frac{9\pi^{4}}{10}$
(b) $\Sigma_{k=5}^{\infty}\frac{1}{(k-2)^{4}}=\Sigma_{k=3}^{\infty}\frac{1}{k^{4}}$
$\Sigma_{k=3}^{\infty}\frac{1}{k^{4}}=\frac{\pi^{4}}{90}-\frac{1}{1^{4}}-\frac{1}{2^{4}}$
$=\frac{\pi^{4}}{90}-\frac{17}{16}$
