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JH

# Euler also found the sum of the $p-$ series with $p = 4:$$\zeta (4) = \displaystyle \sum_{n = 1}^{\infty} \frac {1}{n^4} = \frac {\pi^4}{90}$Use Euler's result to find the sum of the series.(a) $\displaystyle \sum_{n = 1}^{\infty} \left( \frac {3}{n} \right)^4$(b) $\displaystyle \sum_{k = 5}^{\infty} \frac {1}{(k - 2)^4}$

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

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