How do you test for convergence for #sum ln(n)/n^2# for n=1 to infinity?

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Bill K. Share
Jun 1, 2015

Probably the best method is to use the integral test. The function #f(x)=ln(x)/(x^2)# is positive and decreasing for #x\geq 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.

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