Science:Math Exam Resources/Courses/MATH101/April 2012/Question 04 (c)/Solution 1

We will use the integral test. Note that

$\int _{2}^{\infty }{\frac {dx}{x\ln(x)^{101}}}=\lim _{a\to \infty }\int _{2}^{a}{\frac {dx}{x\ln(x)^{101}}}$

We can calculate $\int _{2}^{a}{\frac {dx}{x\ln(x)^{101}}}$ by substitution with $u=\ln x$ (then du=(1/x)dx) to get

${\begin{aligned}\int _{2}^{a}{\frac {dx}{x\ln(x)^{101}}}&=\int _{\ln(2)}^{\ln(a)}u^{-101}du\\&=\left.-{\frac {1}{100u^{100}}}\right|_{\ln(2)}^{\ln(a)}\\&=-{\frac {1}{100}}\left({\frac {1}{(\ln(a))^{100}}}-{\frac {1}{(\ln(2))^{100}}}\right)\end{aligned}}$

Hence

$\int _{2}^{\infty }{\frac {dx}{x\ln(x)^{101}}}=\lim _{a\to \infty }{\frac {1}{100}}\left({\frac {1}{\ln(2)^{100}}}-{\frac {1}{\ln(a)^{100}}}\right)={\frac {1}{100\ln(2)^{100}}}$.

This implies that the series converges absolutely by the integral test.