Probably the best method is to use the integral test. The function f(x)=ln(x)/(x^2)f(x)=ln(x)x2 is positive and decreasing for x\geq 2x≥2. The improper integral \int_{2}^{\infty}f(x)\ dx=lim_{b->\infty}\int_{2}^{b}(ln(x))/(x^2)\ dx can be shown to converge by direct integration (using integration-by-parts with u=ln(x), du=1/x dx, dv=x^[-2} dx, and v=-x^{-1}):
lim_{b->\infty}\int_{2}^{b}(ln(x))/(x^2)\ dx
=lim_{b->\infty}(-ln(b)/b+ln(2)/2)+\int_{2}^{b}x^{-2}\ dx
=ln(2)/2-lim_{b->\infty}((b^{-1}-1/2))=(ln(2)+1)/2.
The integral test now implies that \sum_{n=2}^{\infty}f(n)=\sum_{n=2}^{\infty}\frac{ln(n)}{n^2} converges.
Therefore, \sum_{n=1}^{\infty}\frac{ln(n)}{n^2} converges.
