| Date | May 2013 | Marks available | 3 | Reference code | 13M.3ca.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
Consider the infinite series \(\sum\limits_{n = 1}^\infty {\frac{{{n^2}}}{{{2^n}}}{x^n}} \).
Find the radius of convergence.
Find the interval of convergence.
Given that x = – 0.1, find the sum of the series correct to three significant figures.
Markscheme
\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{{{(n + 1)}^2}{x^{n + 1}}}}{{{2^{n + 1}}}}}}{{\frac{{{n^2}{x^n}}}{{{2^n}}}}}\) M1
\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{{(n + 1)}^2}}}{{{n^2}}} \times \frac{x}{2}\) A1
\( = \frac{x}{2}\) (since \(\lim \to \frac{x}{2}{\text{ as }}n \to \infty \)) A1
the radius of convergence R is found by equating this limit to 1, giving R = 2 A1
[4 marks]
when x = 2, the series is \(\sum {{n^2}} \) which is divergent because the terms do not converge to 0 R1
when x = –2, the series is \(\sum {{{( - 1)}^n}{n^2}} \) which is divergent because the terms do not converge to 0 R1
the interval of convergence is \(] - 2,{\text{ }}2[\) A1
[3 marks]
putting x = – 0.1, (M1)
for any correct partial sum (A1)
– 0.05
– 0.04
– 0.041125
– 0.041025
– 0.0410328 (A1)
the sum is – 0.0410 correct to 3 significant figures A1
[4 marks]
Examiners report
It was pleasing that most candidates were aware of the Radius of Convergence and Interval of Convergence required by parts (a) and (b) of this problem. Many candidates correctly handled the use of the Ratio Test for convergence and there was also the use of Cauchy’s nth root test by a small number of candidates to solve part (a). Candidates need to take care to justify correctly the divergence or convergence of series when finding the Interval of Convergence.
It was pleasing that most candidates were aware of the Radius of Convergence and Interval of Convergence required by parts (a) and (b) of this problem. Many candidates correctly handled the use of the Ratio Test for convergence and there was also the use of Cauchy’s nth root test by a small number of candidates to solve part (a). Candidates need to take care to justify correctly the divergence or convergence of series when finding the Interval of Convergence.
The summation of the series in part (c) was poorly handled by a significant number of candidates, which was surprising on what was expected to be quite a straightforward problem. Again efficient use of the GDC seemed to be a problem. A number of candidates found the correct sum but not to the required accuracy.
