This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 2. Prove that 1 / âˆš 1 + 1 / âˆš 2 + Â·Â·Â· + 1 / âˆš n â‰¥ âˆš n for all n âˆˆ N . We, again, argue using induction. The n = 1 base case follows trivially as both sides are equal to 1 in that case. We then assume that the result holds for n replaced by k1 and seek to use that to prove the n = k case. Here, by this inductive hypothesis, we have 1 / âˆš 1 + 1 / âˆš 2 + Â·Â·Â· + 1 / âˆš k1 + 1 / âˆš k â‰¥ âˆš k1 + 1 âˆš k . Thus, it will suï¬ƒce to show that âˆš k1 + 1 âˆš k â‰¥ âˆš k. Note that p k ( k1) â‰¥ p ( k1) 2 = k1 , for k â‰¥ 2 . Thus, p k ( k1) + 1 â‰¥ k in this range. Or âˆš k1 + 1 âˆš k â‰¥ âˆš k, which is the desired inequality. 1...
View
Full
Document
This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 MetCalfe
 Calculus

Click to edit the document details