Determine whether the series $\sum\limits_{k\,=2}^{\infty }\dfrac{1}{k(\ln k) ^{2}}$ converges or diverges.

Solution The function $f(x)\,{=}\,\dfrac{1}{x (\ln x)^{2}}$ is continuous, positive, and decreasing on the interval $[2,\infty)$. Also $\dfrac{1}{k(\ln k) ^{2}}\,{=}\,f(k)$ for all integers greater than or equal to $2.$ Using the Integral Test, we investigate the improper integral $\int_{2}^{\infty }\dfrac{dx}{x(\ln x)^{2} }.$ We find an antiderivative of $\dfrac{1}{x(\ln x) ^{2}}$ by using the substitution $u=\ln x,$ $du=\dfrac{1}{x}dx.$ Then, $$\int \frac{dx}{x(\ln x)^{2}}=\int \frac{du}{u^{2}}=-\frac{1}{u}+C=-\frac{1 }{\ln x}+C$$

Now we find the improper integral $$\int_{2}^{\infty }\frac{dx}{x(\ln x)^{2}}\!:\,\,\lim_{b\,\rightarrow \,\infty }\,\left[ -\frac{1}{\ln x} \right] _{2}^{b}=\lim_{b\, \rightarrow \,\infty }\,\left( -\frac{1}{\ln b}\right) +\frac{1}{\ln 2}=\frac{1 }{\ln 2}$$

Since the improper integral $\int_{2}^{\infty }\dfrac{dx}{x(\ln x)^{2}}$ converges, the series $\sum\limits_{k\,=2}^{\infty }\dfrac{1}{k(\ln k) ^{2}}$ converges.