Date | November 2008 | Marks available | 11 | Reference code | 08N.3ca.hl.TZ0.2 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Determine | Question number | 2 | Adapted from | N/A |
Question
(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) \(\sqrt {\frac{1}{2}} + \sqrt {\frac{1}{6}} + \sqrt {\frac{1}{{12}}} + \sqrt {\frac{1}{{20}}} + …\)
Markscheme
(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]
Examiners report
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.