Date | None Specimen | Marks available | 9 | Reference code | SPNone.3ca.hl.TZ0.5 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Consider the infinite series \(\sum\limits_{n = 1}^\infty {\frac{{(n - 1){x^n}}}{{{n^2} \times {2^n}}}} \) .
Find the radius of convergence.
Find the interval of convergence.
Markscheme
using the ratio test, \(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{n{x^{n + 1}}}}{{{{(n + 1)}^2}{2^{n + 1}}}} \times \frac{{{n^2}{2^n}}}{{(n - 1){x^n}}}\) M1
\( = \frac{{{n^3}}}{{{{(n + 1)}^2}(n - 1)}} \times \frac{x}{2}\) A1
\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{x}{2}\) A1
the radius of convergence R satisfies
\(\frac{R}{2} = 1\) so R = 2 A1
[4 marks]
considering x = 2 for which the series is
\(\sum\limits_{n = 1}^\infty {\frac{{(n - 1)}}{{{n^2}}}} \)
using the limit comparison test with the harmonic series M1
\(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \), which diverges
consider
\(\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n - 1}}{n} = 1\) A1
the series is therefore divergent for x = 2 A1
when x = –2 , the series is
\(\sum\limits_{n = 1}^\infty {\frac{{(n - 1)}}{{{n^2}}} \times {{( - 1)}^n}} \)
this is an alternating series in which the \({n^{{\text{th}}}}\) term tends to 0 as \(n \to \infty \) A1
consider \(f(x) = \frac{{x - 1}}{{{x^2}}}\) M1
\(f'(x) = \frac{{2 - x}}{{{x^3}}}\) A1
this is negative for \(x > 2\) so the sequence \(\{ |{u_n}|\} \) is eventually decreasing R1
the series therefore converges when x = –2 by the alternating series test R1
the interval of convergence is therefore [–2, 2[ A1
[9 marks]