Apply the Integral Test to determine the convergence of ∑n=2∞1nlnn\sum_{n=2}^{\infty} \frac{1}{n \ln n}∑n=2∞nlnn1
Converges, because f(x)=1xlnxf(x) = \frac{1}{x \ln x}f(x)=xlnx1 is positive and continuous
Diverges, because ∫2∞1xlnxdx=limt→∞ln(lnt)=∞\int_2^{\infty} \frac{1}{x \ln x} dx = \lim_{t \to \infty} \ln(\ln t) = \infty∫2∞xlnx1dx=limt→∞ln(lnt)=∞
The Integral Test is inconclusive for logarithmic functions
Converges, because lnn\ln nlnn grows faster than nnn