Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises - Page 766: 35

(a) $\frac{9\pi^{4}}{10}$ (b) $\frac{\pi^{4}}{90}-\frac{17}{16}$

(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) $\Sigma_{k=5}^{\infty}\frac{1}{(k-2)^{4}}=\Sigma_{k=3}^{\infty}\frac{1}{k^{4}}$ $\Sigma_{k=3}^{\infty}\frac{1}{k^{4}}=\frac{\pi^{4}}{90}-\frac{1}{1^{4}}-\frac{1}{2^{4}}$ $=\frac{\pi^{4}}{90}-\frac{17}{16}$