Find the set of values of k for which the improper integral \(\int_2^\infty {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \) converges.

Show that the series \(\sum\limits_{r = 2}^\infty {\frac{{{{( - 1)}^r}}}{{r\ln r}}} \) is convergent but not absolutely convergent.

consider the limit as \(R \to \infty \) of the (proper) integral

\(\int_2^R {\frac{{{\text{d}}x}}{{x{{(\ln x)}^k}}}} \)     (M1)

substitute \(u = \ln x,{\text{ d}}u = \frac{1}{x}{\text{d}}x\)     (M1)

obtain \(\int_{\ln 2}^{\ln R} {\frac{1}{{{u^k}}}{\text{d}}u = \left[ { - \frac{1}{{k - 1}}\frac{1}{{{u^{k - 1}}}}} \right]_{\ln 2}^{\ln R}} \)     A1 

Note: Ignore incorrect limits or omission of limits at this stage.

or \([\ln u]_{\ln 2}^{\ln R}\) if k = 1     A1 

Note: Ignore incorrect limits or omission of limits at this stage.

because \(\ln R{\text{ }}({\text{and }}\ln \ln R) \to \infty {\text{ as }}R \to \infty \)     (M1)

converges in the limit if k > 1     A1

[6 marks]

C: \({\text{terms}} \to 0{\text{ as }}r \to \infty \)     A1

\(\left| {{u_{r + 1}}} \right| < \left| {{u_r}} \right|\) for all r     A1

convergence by alternating series test     R1

AC: \({(x\ln x)^{ - 1}}\) is positive and decreasing on \([2,\,\infty )\)     A1

not absolutely convergent by integral test using part (a) for k = 1     R1

[5 marks]

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when k = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.

A good number of candidates were able to find the integral in part (a) although the vast majority did not consider separately the integral when k = 1. Many candidates did not explicitly set a limit for the integral to let this limit go to infinity in the anti – derivative and it seemed that some candidates were “substituting for infinity”. This did not always prevent candidates finding a correct final answer but the lack of good technique is a concern. In part (b) many candidates seemed to have some knowledge of the relevant test for convergence but this test was not always rigorously applied. In showing that the series was not absolutely convergent candidates were often not clear in showing that the function being tested had to meet a number of criteria and in so doing lost marks.