Science:Math Exam Resources/Courses/MATH105/April 2015/Question 05 (d)/Solution 1

We will apply the integral test. The test states that if the integral $\int _{n}^{\infty }f(x)\,dx$ converges, the series $\sum _{k=n}^{\infty }f(k)$ converges as well.

We may evaluate the integral $\int _{2}^{\infty }{\frac {1}{x(\ln x)^{3}}}\,dx$ by a substitution. Let $u=\ln x$. Then $du={\frac {dx}{x}}$, which implies that

$\int _{2}^{\infty }{\frac {1}{x(\ln x)^{3}}}\,dx=\int _{\ln 2}^{\infty }{\frac {du}{u^{3}}}=\lim _{b\to \infty }\left.-{\frac {1}{2u^{2}}}\right|_{\ln 2}^{b}=\lim _{b\rightarrow \infty }\left(-{\frac {1}{2b^{2}}}+{\frac {1}{2(\ln 2)^{2}}}\right)={\frac {1}{2(\ln 2)^{2}}}.$

Therefore, the integral $\int _{2}^{\infty }{\frac {1}{x(\ln x)^{3}}}\,dx$ converges, and therefore by the integral test, the series $\sum _{k=2}^{\infty }{\frac {1}{k(\ln k)^{3}}}$ converges also.