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December 20, 2014

December 20, 2014

Posted by **Jenny** on Monday, February 11, 2013 at 6:07pm.

Sigma (lower index n = 4; upper index infinity) 1/(3n^2-2n-15)

- Calculus -
**Steve**, Monday, February 11, 2013 at 6:15pm3n^2-2n-15 > n^2 for n>=4,

the series converges since 1/n^2 converges

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