exam1 - Exam 1 Solutions 1. Claim(1). an an+1 for all n 1....

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Exam 1 Solutions 1. Claim(1). a n a n +1 for all n 1. Proof. By induction on n . Since a 1 = 2 < p 2 + 2 = a 2 the base step is valid. Now assume that a m a m +1 for some m 1. Then a m +1 = 2 + a m p 2 + a m +1 = a m +2 so the induction step is also valid. Claim(2). a n < 2 for all n 1. Proof. By induction on n . Since a 1 = 2 < 2 the base step is valid. Assume that a m < 2 for some m 1. Then a m +1 = 2 + a m < 2 + 2 = 2 so the induction step is also valid. By Claims (1) and (2), we know that ( a n ) is monotone increasing and bounded above by 2. Therefore a n L for some L R by the Monotone Convergence Theorem. Hence, by properties of limits, we have L = lim a n = lim a n +1 = lim 2 + a n = p 2 + lim a n = 2 + L and so L 2 - L - 2 = 0. It follows that L = 2. 2. Suppose a n L and a n L 0 . Given ± > 0 there are N 1 ,N 2 Z such that | a n - L | < ± 2 for all n N 1 and | a n - L 0 | < ± 2 for all n N 2 . So for any n max { N 1 ,N 2 } we have | L - L 0 | = | L - a n + a n - L 0 | ≤ | L - a n | + | a n - L 0 | < ± 2 + ± 2 = ±. Since ± was arbitrary, it follows that | L - L 0 | = 0. Hence L = L 0 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/09/2010 for the course MATH 312 taught by Professor Wynne during the Fall '09 term at Simons Rock.

Page1 / 3

exam1 - Exam 1 Solutions 1. Claim(1). an an+1 for all n 1....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online