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 $,

$ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots \infty= \beta $,

then $ \frac{\alpha}{\beta} $ is equal to :

<p>$$\begin{aligned} & \text { If } \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . \infty=\frac{\pi^4}{90} \quad\text{....... (i)}\\ & \beta=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots, \\ & =\frac{1}{16}\left[\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . .\right], \end{aligned}$$</p> <p>$=\frac{1}{16} \times \frac{\pi^4}{90}$ using (ii) ............ (ii)</p> <p>$$\begin{aligned} & \alpha=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots . . \infty \\ & \left(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots . .\right) \\ & -\left(\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots . .\right) \\ & \alpha=\frac{\pi^4}{90}-\frac{1}{16} \times \frac{\pi^4}{90} \quad \text { [using (i) and (ii)] } \\ & \alpha=\frac{16-1}{16 \times 90} \times \pi^4=\frac{15}{16 \times 90} \pi^4=\frac{\pi^4}{96} \\ & \therefore \frac{\alpha}{\beta}=\frac{\frac{\pi^4}{96}}{\frac{\pi^4}{16 \times 90}}=\frac{16 \times 90}{96}=15 \end{aligned}$$</p>