Animated Solution for Mathematics - Sequence and Series: If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\dots \infty =\frac{{\pi }^4}{90}$, $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\dots \infty =\alpha$, $\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\dots \infty =\beta$, then $\frac{\alpha }{\beta }$ is equal to

Infinite series often appear as daunting, monolithic structures. When you first look at the series:

It is easy to feel overwhelmed by the sheer number of terms. However, the secret to mastering these problems in JEE Advanced is not to calculate the sum, but to understand its internal architecture.

Think of this series as a grand parade of numbers, where every term has a specific position. Our goal is to reorganize this parade.

Odd vs. Even

The problem invites us to partition this series into two distinct sub-series. We define $\alpha$ as the sum of all terms where the denominator is an odd number:

Then, we define $\beta$ as the sum of all terms where the denominator is an even number:

By definition, if we combine these two groups, we reconstruct the original series. Thus, we have the fundamental relationship: $S=\alpha +\beta$. This is our anchor.

Now, let us turn our attention to $\beta$. The even terms have a hidden symmetry. Every denominator is of the form $(2k{)}^4$, where $k$ is a natural number.

Using the exponent law $(ab{)}^n=a^n{b}^n$, we can rewrite the general term as:

Imagine pulling a factor of $\frac{1}{16}$ out of every single term in the $\beta$ series. When we do this, the series transforms into:

Suddenly, the expression inside the parentheses is identical to our original series $S$. We have successfully established that $\beta =\frac{1}{16}S$. This is the pivotal moment where the complexity collapses into simplicity.

With $\beta$ expressed in terms of $S$, finding $\alpha$ becomes trivial. We return to our anchor equation $S=\alpha +\beta$.

Substituting our finding for $\beta$, we get $S=\alpha +\frac{1}{16}S$. Rearranging this, we find:

Now, we simply calculate the ratio $\frac{\alpha }{\beta }$:

The $S$ terms cancel, the $16$s cancel, and we are left with $15$. The elegance of this result is a testament to the power of algebraic manipulation over brute-force calculation.

A Generalization

Before we conclude, let us look at the bigger picture. This problem is a specific instance of a broader rule.

For any series of the form $S={\sum }_{n=1}^{\infty }\frac{1}{n^p}$, the ratio of the sum of odd terms to the sum of even terms is always $2^p-1$.

When $p=4$, we get $2^4-1=15$. If the power were $2$, the ratio would be $2^2-1=3$.

This is the kind of insight that separates the top rankers from the rest. Keep looking for these patterns, and you will find that even the most complex problems have a simple, beautiful core.