Use the integral test to determine whether or not \(\sum\limits_{n = 2}^\infty  {\frac{1}{{n{{\left( {{\text{ln}}\,n} \right)}^2}}}} \) converges.

let \(u = {\text{ln}}\,x\)     (M1)

\( \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}\)

\(\int {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \int {\frac{1}{{{u^2}}}} \,{\text{d}}u\)      (A1)

\( =  - \frac{1}{u} =  - \frac{1}{{{\text{ln}}\,x}}\)      (A1)

\(\int\limits_2^m {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \left[ { - \frac{1}{{{\text{ln}}\,x}}} \right]_2^m\)     M1

\( = \left[ { - \frac{1}{{{\text{ln}}\,m}} +  - \frac{1}{{{\text{ln}}\,2}}} \right]\)     A1

as \(m \to \infty \), \( - \frac{1}{{{\text{ln}}\,m}} \to 0\)      (A1)

\(\int\limits_2^\infty  {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \frac{1}{{{\text{ln}}\,2}}\) and hence the series converges       R1

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