It is given that _n=1^(1)/(n^4) = (^4)/(90).

Question

It is given that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}$. If $\alpha = \dfrac{1}{1^4}+\dfrac{1}{3^4}+\dfrac{1}{5^4}+\cdots\infty$ and $\beta = \dfrac{1}{2^4}+\dfrac{1}{4^4}+\dfrac{1}{6^4}+\cdots\infty$, then $\dfrac{\alpha}{\beta}$ is equal to:

Answer: D (15)

Accepted Answer

$$\beta = \frac{1}{2^4}\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{1}{16}\cdot\frac{\pi^4}{90}.$$

$$\alpha = \frac{\pi^4}{90} - \beta = \frac{\pi^4}{90}\left(1-\frac{1}{16}\right) = \frac{15}{16}\cdot\frac{\pi^4}{90} = \frac{\pi^4}{96}.$$

$$\frac{\alpha}{\beta} = \frac{\pi^4/96}{\pi^4/(16\times90)} = \frac{16\times90}{96} = 15.$$

Solutions · 1