Asked by Jenny
Test for convergence or divergence. Indicate the test that was used and justify your answer.
Sigma (lower index n = 4; upper index infinity) 1/(3n^2-2n-15)
Sigma (lower index n = 4; upper index infinity) 1/(3n^2-2n-15)
Answers
Answered by
Steve
3n^2-2n-15 > n^2 for n>=4,
the series converges since 1/n^2 converges
the series converges since 1/n^2 converges
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