Determine whether the series \sum_{n=1}^{\infty} \frac{\ln(2n)}{3n} converges or diverges.
Solution
The function $f(x) = \frac{\ln(2x)}{3x}$ is positive and continuous for $x >$ because the logarithm function is continuous. But it is not obvious whether or not $f$ is decreasing, so we compute its derivative.
$f'(x) = \frac{\frac{1}{x} \times 3x - 3 \ln(2x)}{9x^2} = \frac{1 - \ln(2x)}{3x^2}$
Thus $f'(x) < 0$ when $\ln(2x) > 0$, that is, $x > \ln(2x) > 1$. It follows that $f$ is decreasing when $x > e$
and so we can apply the Integral Test:
$\int_{1}^{\infty} \frac{\ln(2x)}{3x} dx = \lim_{t \to \infty} \int_{1}^{t} \frac{\ln(2x)}{3x} dx$
$= \lim_{t \to \infty} \left[ \frac{(\ln(2t))^2}{6} - \frac{(\ln(2))^2}{6} \right] = \infty$.
Since this improper integral is divergent, the series \sum_{n=1}^{\infty} \frac{\ln(2n)}{3n} is also divergent by the Integral Test.
