Trail to infinity

We are given that $\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^{2}}=S$ .

$=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\ldots \infty$

$=\underbrace{\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^2}}_{S}-\underbrace{\displaystyle \sum^{\infty}_{n=1}\frac{1}{(2n)^2}}_{\frac{S}{4}}$

$=S-\frac{S}{4}=\boxed{\frac{3S}{4}}$

An intuitive solution can be deduced by eliminating the incorrect answers to find the correct one. The first term of the infinite sum is 1, and all other terms are decimals below 1 + 0.25 + 0.11 + 0.0625 + 0.04 + ...., as the sum approaches 2. Since the first term is also 1 for the second series $\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2}$ , then this sum will also be above 1. Thus the numerator and denominator of the fraction of S will be $\frac{1.x}{1.y}S$ , where x and y will be numbers from the sum of all other terms after the first term for the first and second series respectively. Therefore by elimination, the fraction cannot be $\frac{1}{4}$ , $\frac{1}{3}$ or $\frac{1}{2}$ , it can only be between $\frac{1}{2}$ and 1. $\newline$ Thus the answer has to be $\frac{3}{4}S$ .