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]