# calculus

test the series for convergence or divergence.
the sum from n=1 to infinity of ((-1)^n*e^n)/(n^3)
I said it converges because the derivative of (1/n^3) is decreasing

1. Hi:

The series is alternating of the form(a_n)*(-1)^n, where a_n = e^n/n^3. Because limit (n->inf)[a_n] not= 0 (a_n->inf as n->inf}, the series diverges.

Regards,

Rich B.

posted by rich

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