This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: } , either n is even or odd, and in either case  s ns  < Â± . For the rest of the exercise, let s n = 11 2 + 1 3 Â·Â·Â· + (1) n +1 1 n . (b) (5 pts) Prove that ( s 2 n ) is increasing and bounded above. Prove that ( s 2 n +1 ) is decreasing and bounded below. Solution: s 2( n +1)s 2 n = (1) 2 n +3 1 2 n + 2 +(1) 2 n +2 1 2 n + 1 =1 2 n + 2 + 1 2 n + 1 > so the subsequence is increasing. It is bounded above by 1 because s 2 n = 11 2 + 1 3Â·Â·Â· +(1) 2 n +1 1 2 n = 1Â± 1 21 3 Â²Â·Â·Â·Â± 1 2 n21 2 n1 Â²1 2 n , and all terms in the parentheses are positive. Proof that ( s 2 n +1 ) is decreasing and bounded below by 0 is similar. (c) (2 pts) Prove that lim s 2 n = lim s 2 n +1 . Conclude ( s n ) converges. Solution: lim s 2 n = lim s 2 n +11 2 n +1 = lim s 2 n +1 because lim 1 2 n +1 = 0....
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.
 Winter '11
 Wiley
 Differential Equations, Equations

Click to edit the document details