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 :

The given infinite series $$\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots$$ equals $$\frac{\pi^{4}}{90}$$. This is the value of the Riemann zeta function $$\zeta(4)$$, so we may write

$$\zeta(4)=\sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{\pi^{4}}{90}\;.-(1)$$

Split the same sum into its odd-index and even-index parts:

$$\zeta(4)=\Bigl(\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\ldots\Bigr)+\Bigl(\frac{1}{2^{4}}+\frac{1}{4^{4}}+\frac{1}{6^{4}}+\ldots\Bigr) =\alpha+\beta\;.-(2)$$

First evaluate $$\beta$$, the sum over even integers. Write each even number as $$2k$$, where $$k=1,2,3,\ldots$$:

$$\beta=\sum_{k=1}^{\infty}\frac{1}{(2k)^{4}} =\sum_{k=1}^{\infty}\frac{1}{2^{4}}\cdot\frac{1}{k^{4}} =\frac{1}{16}\sum_{k=1}^{\infty}\frac{1}{k^{4}} =\frac{1}{16}\,\zeta(4)\;.-(3)$$

Substitute $$\zeta(4)=\frac{\pi^{4}}{90}$$ from $$(1)$$ into $$(3)$$:

$$\beta=\frac{1}{16}\cdot\frac{\pi^{4}}{90} =\frac{\pi^{4}}{1440}\;.-(4)$$

Now find $$\alpha$$ using $$(2)$$:

$$\alpha=\zeta(4)-\beta =\frac{\pi^{4}}{90}-\frac{\pi^{4}}{1440}\;.-(5)$$

Express both fractions in $$(5)$$ with the common denominator $$1440$$:

$$\frac{\pi^{4}}{90}=\frac{16\pi^{4}}{1440}\,,\quad \alpha=\frac{16\pi^{4}}{1440}-\frac{\pi^{4}}{1440} =\frac{15\pi^{4}}{1440} =\frac{\pi^{4}}{96}\;.-(6)$$

Finally, compute the required ratio:

$$\frac{\alpha}{\beta} =\frac{\pi^{4}/96}{\pi^{4}/1440} =\frac{1440}{96} =15\;.-(7)$$

Therefore, $$\frac{\alpha}{\beta}=15$$. Hence the correct option is Option C.