Prove :

newtoinequality · #1 $Old$ 11-01-2008, 11:44 AM

Show that for any natural number the sum

$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots + \frac{1}{n^2}$

lies between the values

$\left(1-\frac{2}{n+1}\right)\left(1-\frac{2}{2n+1}\right)\frac{\pi^2}{6}$ and $\left(1-\frac{1}{2n+1}\right)\left(1+\frac{1}{2n+1}\right)\frac{\pi^2}{6}.$

EyesForEars · #2 $Old$ 11-17-2008, 07:07 AM

To start this, would one have to find the domain of the these two expressions over the natural numbers?

and

ThePerfectHacker · #3 $Old$ 11-17-2008, 07:44 AM

Quote:

Originally Posted by newtoinequality $View Post$

Show that for any natural number the sum

$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots + \frac{1}{n^2}$

lies between the values

$\left(1-\frac{2}{n+1}\right)\left(1-\frac{2}{2n+1}\right)\frac{\pi^2}{6}$ and $\left(1-\frac{1}{2n+1}\right)\left(1+\frac{1}{2n+1}\right)\frac{\pi^2}{6}.$

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