User:Sethdoe92~simplewiki

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Wiki

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neoi (?!?!)

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yo

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mama

For $\alpha >0$ aa

$\sum _{n=1}^{\infty }{\frac {1}{n^{\alpha }}}<\infty \ \Leftrightarrow \ \alpha >1$

$\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}$

blah blah test

TESTING TESTING WOOP WOOP WOOP

411 This user strives to maintain a policy of neutrality on controversial issues. MThis Article has Been Rated MATURE for Being the highest level of disturbing content.