If frac{1}{1^4}+frac{1}{2^4}+frac{1}{3^4}+ldo | vedclass

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(C) Given $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots = \frac{\pi^4}{90}$.$\beta = \frac{1}{2^4}+\frac{1}{4^

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) Given $\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots = \frac{\pi^4}{90}$.$\beta = \frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots = \frac{1}{2^4} \left( \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots ight) = \frac{1}{16} 	imes \frac{\pi^4}{90}$.$\alpha = \frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots = \left( \sum_{n=1}^{\infty} \frac{1}{n^4} ight) - \beta = \frac{\pi^4}{90} - \frac{1}{16} 	imes \frac{\pi^4}{90} = \frac{15}{16} 	imes \frac{\pi^4}{90}$.Therefore,$\frac{\alpha}{\beta} = \frac{\frac{15}{16} 	imes \frac{\pi^4}{90}}{\frac{1}{16} 	imes \frac{\pi^4}{90}} = 15$.

If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots \infty = \frac{\pi^4}{90}$,$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots \infty = \alpha$,and $\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots \infty = \beta$,then $\frac{\alpha}{\beta}$ is equal to:

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