View source for 2013 PMWC Problems/Problem I15

==Solution==

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

Splitting <math>M</math> into its even component and odd component, we can write <math>M</math> as:
<cmath>\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*}</cmath>

Simplifying, we see that
<cmath>\begin{align*}
\frac{3}{4}M&=K\\
\frac{M}{K}&=\frac{4}{3}
\end{align*}</cmath>

Therefore, the required answer is <math>\boxed{\frac{4}{3}}</math>