CAIE
Further Paper 1
2020
November
— Question 3
1 marks
Exam Board
CAIE
Module
Further Paper 1 (Further Paper 1)
Year
2020
Session
November
Marks
1
Topic
Sequences and series, recurrence and convergence
3
By simplifying \(\left( x ^ { n } - \sqrt { x ^ { 2 n } + 1 } \right) \left( x ^ { n } + \sqrt { x ^ { 2 n } + 1 } \right)\), show that \(\frac { 1 } { x ^ { n } - \sqrt { x ^ { 2 n } + 1 } } = - x ^ { n } - \sqrt { x ^ { 2 n } + 1 }\). [1]
Let \(u _ { n } = x ^ { n + 1 } + \sqrt { x ^ { 2 n + 2 } + 1 } + \frac { 1 } { x ^ { n } - \sqrt { x ^ { 2 n } + 1 } }\).
Use the method of differences to find \(\sum _ { \mathrm { n } = 1 } ^ { \mathrm { N } } \mathrm { u } _ { \mathrm { n } }\) in terms of \(N\) and \(x\).
Deduce the set of values of \(x\) for which the infinite series
$$u _ { 1 } + u _ { 2 } + u _ { 3 } + \ldots$$
is convergent and give the sum to infinity when this exists.