IB DP Maths Topic 9.2 Convergence of infinite series HL Paper 3

(a)     \(\sum\limits_{n = 0}^\infty  {\left( {\sin \frac{{n\pi }}{2} – \sin \frac{{(n + 1)\pi }}{2}} \right)} \)

\( = \left( {\sin 0 – \sin \frac{\pi }{2}} \right) + \left( {\sin \frac{\pi }{2} – \sin \pi } \right) + \left( {\sin \pi  – \sin \frac{{3\pi }}{2}} \right) + \left( {\sin \frac{{3\pi }}{2} – \sin 2\pi } \right) + \ldots \)     (M1)

the \({n^{{\text{th}}}}\) term is ±1 for all n, i.e. the \({n^{{\text{th}}}}\) term does not tend to 0     A1

hence the series does not converge     A1

[3 marks]

(b)     EITHER

using the ratio test     (M1)

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\text{e}}^{n + 1}}}}{{{\pi ^{n + 1}}}}} \right)\left( {\frac{{{\pi ^n}}}{{{{\text{e}}^n} – 1}}} \right)\)     M1A1

\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\text{e}}^{n + 1}} – 1}}{{{{\text{e}}^n} – 1}}} \right)\left( {\frac{{{\pi ^n}}}{{{\pi ^{n + 1}}}}} \right) = \frac{{\text{e}}}{\pi }\,\,\,\,\,( \approx 0.865)\)     M1A1

\(\frac{{\text{e}}}{\pi } < 1\), hence the series converges     R1A1

OR

\(\sum\limits_{n = 1}^\infty  {\frac{{{{\text{e}}^n} – 1}}{{{\pi ^n}}} = \sum\limits_{n = 1}^\infty  {{{\left( {\frac{{\text{e}}}{\pi }} \right)}^n} – {{\left( {\frac{1}{\pi }} \right)}^n} = \sum\limits_{n = 1}^\infty  {{{\left( {\frac{{\text{e}}}{\pi }} \right)}^n} – \sum\limits_{n = 1}^\infty  {{{\left( {\frac{{\text{1}}}{\pi }} \right)}^n}} } } } \)     M1A1

the series is the difference of two geometric series, with \(r = \frac{{\text{e}}}{\pi }\,\,\,\,\,( \approx 0.865)\)     M1A1

and \(\frac{1}{\pi }\,\,\,\,\,( \approx 0.318)\)     A1

for both \(\left| r \right| < 1\), hence the series converges     R1A1

OR

\(\forall n,{\text{ }}0 < \frac{{{{\text{e}}^n} – 1}}{{{\pi ^n}}} < \frac{{{{\text{e}}^n}}}{{{\pi ^n}}}\)     (M1)A1A1

the series \(\frac{{{{\text{e}}^n}}}{{{\pi ^n}}}\) converges since it is a geometric series such that \(\left| r \right| < 1\)     A1R1

therefore, by the comparison test, \(\frac{{{{\text{e}}^n} – 1}}{{{\pi ^n}}}\) converges     R1A1

[7 marks]

(c)     by limit comparison test with \(\frac{{\sqrt n }}{{{n^2}}}\),     (M1)

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

hence both series converge or both diverge     R1

by the p-test \(\sum\limits_{n = 1}^\infty  {\frac{{\sqrt n }}{{{n^2}}} = {n^{\frac{{ – 3}}{2}}}} \) converges, hence both converge     R1A1

[6 marks]

Total [16 marks]

This was the least successfully answered question on the paper. Candidates often did not know which convergence test to use; hence very few full successful solutions were seen. The communication of the method used was often quite poor.

a) Many candidates failed to see that this is a telescoping series. If this was recognized then the question was fairly straightforward. Often candidates unsuccessfully attempted to apply the standard convergence tests.

b) Many candidates used the ratio test, but some had difficulty in simplifying the expression. Others recognized that the series is the difference of two geometric series, and although the algebraic work was done correctly, some failed to communicate the conclusion that since the absolute value of the ratios are less than 1, hence the series converges. Some candidates successfully used the comparison test.

c) Although the limit comparison test was attempted by most candidates, it often failed through an inappropriate selection of a series.