# Forget the potatoes, have some pie!

Level pendingSasha has recently stopped eating potatoes and switched to eating pie, particularly enjoying apple pie (much to Jean's delight). When Levi went off on one of his rants about a topic no one cares about, he started talking about the Riemann zeta function, where \[\zeta(k)=\sum_{n=1}^\infty\dfrac{1}{n^k}\] Her ears perk up when Levi says that \(\zeta(2)=\frac{\pi^2}{6}\) and \(\zeta(4)=\frac{\pi^4}{90}.\) She wants more of this "zeta function" and its pie. So she tries to find something along the lines of \[\sum_{k=1}^\infty\zeta(k)\] She quickly realizes this is irrational, but that it won't be if you remove the \(1^k\) term from the expansion of \(\zeta(k).\) In addition, values of \(k\) for which the value of \(\zeta(k)\) has \(\pi\) somewhere in them only occur when \(k\) is even.

Sasha was wowed to see that \(\displaystyle\sum_{k=1}^\infty(\zeta(2k)-1)=T.\) What is \(\lfloor100T\rfloor?\)

\(\textbf{Details and Assumptions}\)

The first few terms of Sasha's series are \(\left(\frac{\pi^2}{6}-1\right)+\left(\frac{\pi^4}{90}-1\right)+\left(\frac{\pi^6}{945}-1\right)+\left(\frac{\pi^8}{9450}-1\right)+\ldots\)

**Your answer seems reasonable.**Find out if you're right!

**That seems reasonable.**Find out if you're right!

Already have an account? Log in here.