Date | November 2011 | Marks available | 4 | Reference code | 11N.3ca.hl.TZ0.4 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Show | Question number | 4 | Adapted from | N/A |
Question
Using the integral test, show that ∞∑n=114n2+1 is convergent.
(i) Show, by means of a diagram, that ∞∑n=114n2+1<14×12+1+∫∞114x2+1dx.
(ii) Hence find an upper bound for ∞∑n=114n2+1
Markscheme
∫14x2+1dx=12arctan2x+k (M1)(A1)
Note: Do not penalize the absence of “+k”.
∫∞114x2+1dx=12lima→∞[arctan2x]a1 (M1)
Note: Accept 12[arctan2x]∞1.
=12(π2−arctan2)(=0.232) A1
hence the series converges AG
[4 marks]
(i)
A2
The shaded rectangles lie within the area below the graph so that ∞∑n=214n2+1<∫∞114x2+1dx. Adding the first term in the series, 14×12+1, gives ∞∑n=114n2+1<14×12+1+∫∞114x2+1dx R1AG
(ii) upper bound =15+12(π2−arctan2)(=0.432) A1
[4 marks]
Examiners report
This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate 14x2+1 successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.
This proved to be a hard question for most candidates. A number of fully correct answers to (a) were seen, but a significant number were unable to integrate 14x2+1 successfully. Part (b) was found the hardest by candidates with most candidates unable to draw a relevant diagram, without which the proof of the inequality was virtually impossible.