(i)     Show that \(\int_1^\infty  {\frac{1}{{x(x + p)}}{\text{d}}x,{\text{ }}p \ne 0} \) is convergent if p > −1 and find its value in terms of p.

(ii)     Hence show that the following series is convergent.

\[\frac{1}{{1 \times 0.5}} + \frac{1}{{2 \times 1.5}} + \frac{1}{{3 \times 2.5}} + ...\]

Determine, for each of the following series, whether it is convergent or divergent.

(i)     \(\sum\limits_{n = 1}^\infty  {\sin \left( {\frac{1}{{n(n + 3)}}} \right)} \)

(ii)     \(\sqrt {\frac{1}{2}}  + \sqrt {\frac{1}{6}}  + \sqrt {\frac{1}{{12}}}  + \sqrt {\frac{1}{{20}}}  + …\)

(i)     the integrand is non-singular on the domain if p > –1 with the latter assumed, consider

\(\int_1^R {\frac{1}{{x(x + p)}}} {\text{d}}x = \frac{1}{p}\int_1^R {\frac{1}{x} - \frac{1}{{x + p}}{\text{d}}x} \)     M1A1

\( = \frac{1}{p}\left[ {\ln \left( {\frac{x}{{x + p}}} \right)} \right]_1^R,{\text{ }}p \ne 0\)     A1

this evaluates to

\(\frac{1}{p}\left( {\ln \frac{R}{{R + p}} - \ln \frac{1}{{1 + p}}} \right),{\text{ }}p \ne 0\)     M1

\( \to \frac{1}{p}\ln (1 + p)\)     A1

because \(\frac{R}{{R + p}} \to 1{\text{ as }}R \to \infty \)     R1

hence the integral is convergent     AG

(ii)     the given series is \(\sum\limits_{n = 1}^\infty  {f(n),{\text{ }}f(n) = \frac{1}{{n(n - 0.5)}}} \)     M1

the integral test and p = –0.5 in (i) establishes the convergence of the series     R1

[8 marks]

(i)     as we have a series of positive terms we can apply the comparison test, limit form

comparing with \(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \)     M1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {\frac{1}{{n(n + 3)}}} \right)}}{{\frac{1}{{{n^2}}}}} = 1\)     M1A1

as \(\sin \theta \approx \theta {\text{ for small }}\theta \)     R1

and \(\frac{{{n^2}}}{{n(n + 3)}} \to 1\)     R1

(so as the limit (of 1) is finite and non-zero, both series exhibit the same behavior)

\(\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \) converges, so this series converges     R1

(ii)     the general term is

\(\sqrt {\frac{1}{{n(n + 1)}}} \)     A1

\(\sqrt {\frac{1}{{n(n + 1)}}}  > \sqrt {\frac{1}{{(n + 1)(n + 1)}}} \)     M1

\(\sqrt {\frac{1}{{(n + 1)(n + 1)}}}  = \frac{1}{{n + 1}}\)     A1

the harmonic series diverges     R1

so by the comparison test so does the given series     R1

[11 marks]

Part(a)(i) caused problems for some candidates who failed to realize that the integral can only be tackled by the use of partial fractions. Even then, the improper integral only exists as a limit – too many candidates ignored or skated over this important point. Candidates must realize that in this type of question, rigour is important, and full marks will only be awarded for a full and clearly explained argument. This applies as well to part(b), where it was also noted that some candidates were confusing the convergence of the terms of a series to zero with convergence of the series itself.

Part(a)(i) caused problems for some candidates who failed to realize that the integral can only be tackled by the use of partial fractions. Even then, the improper integral only exists as a limit – too many candidates ignored or skated over this important point. Candidates must realize that in this type of question, rigour is important, and full marks will only be awarded for a full and clearly explained argument. This applies as well to part(b), where it was also noted that some candidates were confusing the convergence of the terms of a series to zero with convergence of the series itself.