Question

Determine whether the series is convergent or divergent.

\(\sum\limits_{{\rm{n = 2}}}^\infty {\frac{{\rm{1}}}{{{\rm{n}}\sqrt {{\rm{lnn}}} }}} \)

Short answer

The series is divergent.

Step-by-step solution

Integral test.

Let \({\rm{f(n) &= }}\frac{{\rm{1}}}{{{\rm{n}}\sqrt {{\rm{lnn}}} }}{\rm{.f(n)}}\)is continuous, positive.

Decreasing on \({\rm{(2,}}\infty {\rm{)}}\)

\begin{aligned}\int_{\rm{2}}^\infty {\rm{f}} {\rm{(x)dx &= }}\int_{\rm{2}}^\infty {\frac{{\rm{1}}}{{{\rm{x}}\sqrt {{\rm{lnx}}} }}} {\rm{dx}}\\{\rm{ &= }}\int_{\rm{2}}^\infty {\frac{{\rm{1}}}{{\sqrt {{\rm{lnx}}} }}} {\rm{d(lnx)}}\\{\rm{ &= }}\left. {{\rm{2}}\sqrt {{\rm{lnx}}} } \right|_{\rm{2}}^\infty \\{\rm{ &= }}\infty \end{aligned}

Result.

\(\int_{\rm{2}}^\infty {\rm{f}} {\rm{(x)dx = }}\infty \)

As a result, the series is a divergent one.