finalsolnmat25

# finalsolnmat25 - Final Solutions 1. Calculations. Find the...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Final Solutions 1. Calculations. Find the following quantities. No proof required, but show any sidework for partial credit. (a) (3 pts) X n =2 1 2 n . Solution: X n =2 1 2 n = X n =0 1 2 n- 1- 1 2 = 1 1- 1 2- 1- 1 2 = 1 2 . (b) (4 pts) The limit of the sequence 2 , p 2 2 , q 2 p 2 2 ,... . Solution: The sequence can be written recursively as a n +1 = 2 a n . So, assuming ( a n ) converges, letting a = lim a n , we get the equation a = 2 a . The solutions are a = 0 or a = 2. Because the sequence is increasing and starts out positive, the limit must be a = 2. (c) (3 pts) The value lim sup s n where s n = (- 1) n 4 n- 3 3 n +4 . Solution: The limit of 4 n- 3 3 n +4 is 4 3 , so the sequence ( s n ) eventually oscillates between values close to 4 3 and values close to- 4 3 . Therefore, limsup s n = 4 3 . (d) (2 pts) The set of limit points of S = [0 , 1) { 3 } . The set of limit points is [0 , 1]. Note that 3 is not a limit point, its an isolated point. (e) (3 pts) X n =1 1 n ( n + 1) . This can be written as a telescoping series whose limit is 1. (see homework 5 solutions.) 1 2. Suppose ( a n ) is a sequence of real numbers so that limsup a n = . Assume R . (a) (5 pts) Let &gt; 0 be arbitrary. Prove by contradiction that there exist infinitely many sequence elements in the neighborhood ( - , + )....
View Full Document

## This note was uploaded on 10/16/2011 for the course MATH 34 taught by Professor Wiley during the Winter '11 term at UC Merced.

### Page1 / 7

finalsolnmat25 - Final Solutions 1. Calculations. Find the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online