# 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 $f \left(x\right) = \ln \frac{x}{{x}^{2}}$ is positive and decreasing for $x \setminus \ge q 2$. The improper integral $\setminus {\int}_{2}^{\setminus \infty} f \left(x\right) \setminus \mathrm{dx} = {\lim}_{b \to \setminus \infty} \setminus {\int}_{2}^{b} \frac{\ln \left(x\right)}{{x}^{2}} \setminus \mathrm{dx}$ can be shown to converge by direct integration (using integration-by-parts with $u = \ln \left(x\right)$, $\mathrm{du} = \frac{1}{x} \mathrm{dx}$, $\mathrm{dv} = {x}^{- 2} \mathrm{dx}$, and $v = - {x}^{- 1}$):

${\lim}_{b \to \setminus \infty} \setminus {\int}_{2}^{b} \frac{\ln \left(x\right)}{{x}^{2}} \setminus \mathrm{dx}$

$= {\lim}_{b \to \setminus \infty} \left(- \ln \frac{b}{b} + \ln \frac{2}{2}\right) + \setminus {\int}_{2}^{b} {x}^{- 2} \setminus \mathrm{dx}$

$= \ln \frac{2}{2} - {\lim}_{b \to \setminus \infty} \left(\left({b}^{- 1} - \frac{1}{2}\right)\right) = \frac{\ln \left(2\right) + 1}{2}$.

The integral test now implies that $\setminus {\sum}_{n = 2}^{\setminus \infty} f \left(n\right) = \setminus {\sum}_{n = 2}^{\setminus \infty} \setminus \frac{\ln \left(n\right)}{{n}^{2}}$ converges.

Therefore, $\setminus {\sum}_{n = 1}^{\setminus \infty} \setminus \frac{\ln \left(n\right)}{{n}^{2}}$ converges.

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