Use the integral test to determine whether the infinite series $\sum\limits_{n = 2}^\infty {\frac{1}{{n\sqrt {\ln n} }}}$ is convergent or divergent.

consider $I = \int\limits_2^N {\frac{{{\text{d}}x}}{{x\sqrt {\ln x} }}}$     M1A1

Note:     Do not award A1 if $n$ is used as the variable or if lower limit equal to 1, but some subsequent A marks can still be awarded. Allow $\infty$ as upper limit.

let $y = \ln x$     (M1)

${\text{d}}y = \frac{{{\text{d}}x}}{x},$     (A1)

$\left[ {2,{\text{ }}N} \right] \Rightarrow \left[ {\ln 2,{\text{ }}\ln N} \right]$

$I = \int\limits_{\ln \,2}^{\ln \,N} {\frac{{{\text{d}}y}}{{\sqrt y }}}$    (A1)

Note:     Condone absence of limits, or wrong limits.

$= \left[ {2\sqrt y } \right]_{\ln 2}^{\ln N}$     A1

Note:     A1 is for the correct integral, irrespective of the limits used. Accept correct use of integration by parts.

$= 2\sqrt {\ln N} - 2\sqrt {\ln 2}$     (M1)

Note:     M1 is for substituting their limits into their integral and subtracting.

$\to \infty {\text{ as }}N \to \infty$     A1

Notes:     Allow “$= \infty$”, “limit does not exist”, “diverges” or equivalent.

Do not award if wrong limits substituted into the integral but allow $N$ or $\infty$ as an upper limit in place of $\ln N$.

(by the integral test) the series is divergent (because the integral is divergent)     A1

Notes:     Do not award this mark if $\infty$ used as upper limit throughout.

[9 marks]