Forget the potatoes, have some pie!

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Sasha 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 ζ(k)=n=11nk\zeta(k)=\sum_{n=1}^\infty\dfrac{1}{n^k} Her ears perk up when Levi says that ζ(2)=π26\zeta(2)=\frac{\pi^2}{6} and ζ(4)=π490.\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 k=1ζ(k)\sum_{k=1}^\infty\zeta(k) She quickly realizes this is irrational, but that it won't be if you remove the 1k1^k term from the expansion of ζ(k).\zeta(k). In addition, values of kk for which the value of ζ(k)\zeta(k) has π\pi somewhere in them only occur when kk is even.

Sasha was wowed to see that k=1(ζ(2k)1)=T.\displaystyle\sum_{k=1}^\infty(\zeta(2k)-1)=T. What is 100T?\lfloor100T\rfloor?

Details and Assumptions\textbf{Details and Assumptions}

The first few terms of Sasha's series are (π261)+(π4901)+(π69451)+(π894501)+\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


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