1 oh probability and markov chains math 25612571 e09

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 1/2571 E09 Proof. We consider the sequences An = def n! nn+1/2 e-n , an = log An def and show that an decreases but an - 1/(12n) increases as n . This implies that limn an = a 0 and thus limn An = A = ea 1. By a straightforward computation, log 2n + 1 n+1 2n + 1 1 + (2n + 1)-1 An = log -1 = log - 1, An+1 2 n 2 1 - (2n + 1)-1 1+x so it is enough to estimate above the function f (x) = 1 log 1-x for |x| < 1. To 2 this end, define 1 1+x x3 g(x) = log -x- . 2 1-x 3(1 - x2 ) We have x2 1 1 2x4 2x4 1 -1- + - =- 0 g...
View Full Document

Ask a homework question - tutors are online