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[Solution] Sum of reciprocal squares.

Problem was


1) Show that π4>22

2) Show that n=21n2>35


For part 1) we can give a geometric proof of sorts.


















A circle of radius 1. The length of arc AB is π4. The length of the line segment AB is 2sinπ8=22. Since arc length > line segment length, we are done.




For part 2)


We use the following inequality

n2<(n13)(n+23)



This gives us

1n2>1n131n+23


Setting n=2,3, and adding we get a telescoping series on the right and the only term that remains is 1213=35.

This implies part 1 because of the classic result that n=11n2=π26.


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