Calculus BC: Series

Letting u = log(x) , we compute the relevant integral:

The limit of this integral as m→∞ exists and is equal to 1/log(2) , so the sum converges by the integral test.

Letting u = x ³ + 1 , we compute the relevant integral:

This quantity clearly has no limit as m→∞ , so the series diverges. Notice that there are far more efficient ways to show the divergence of this series; for instance one could use the comparison test with the harmonic series 1/n .