2013 PMWC Problems/Problem I15

Solution

First of all, note that $M=\sum_{n=1}^\infty \frac{1}{k^2}$ , and that $K=\sum_{n=1}^\infty \frac{1}{(2k-1)^2}$

Splitting $M$ into its even component and odd component, we can write $M$ as: $\begin{align*} M&=\sum_{k=1}^\infty \frac{1}{k^2}\\ &=\sum_{k=1}^\infty \frac{1}{(2k)^2} + \frac{1}{(2k-1)^2}\\ &=\sum_{k=1}^\infty \frac{1}{(2k)^2} + \sum_{k=1}^\infty \frac{1}{(2k-1)^2}\\ &=K+ \sum_{k=1}^\infty \frac{1}{4k^2}\\ &=K+ \frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^2}\\ &=K+\frac{1}{4}M\\ \end{align*}$

Simplifying, we see that $\begin{align*} \frac{3}{4}M&=K\\ \frac{M}{K}&=\frac{4}{3} \end{align*}$

Therefore, the required answer is $\boxed{\frac{4}{3}}$

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2013 PMWC Problems/Problem I15

Solution