Using the Integral Test, determine the convergence of ∑n=2∞1n(lnn)2\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}∑n=2∞n(lnn)21.
Converges; ∫2∞1x(lnx)2dx\int_2^{\infty} \frac{1}{x(\ln x)^2} dx∫2∞x(lnx)21dx converges
Diverges; ∫2∞1x(lnx)2dx\int_2^{\infty} \frac{1}{x(\ln x)^2} dx∫2∞x(lnx)21dx diverges
The Integral Test does not apply because lnx\ln xlnx is not polynomial
Converges; by comparison to ∑1n2\sum \frac{1}{n^2}∑n21