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March 28, 2015

March 28, 2015

Posted by **Jenny** on Monday, February 11, 2013 at 6:06pm.

Sigma (lower index n = 1; upper index infinity) [sin((2n-1)pi/2)]/n

A. The series diverges

B. The series converges conditionally.

C. The series converges absolutely.

D. It cannot be determined.

- Calculus -
**Steve**, Monday, February 11, 2013 at 6:12pmsince sin(nth odd multiple of pi/2) = -(-1)^n the sequence is

1 - 1/2 + 1/3 - 1/4 +...

The alternating harmonic series converges to ln(2)

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