let an be a sequence of nonzero numbers Prove that if limsup

let an be a sequence of nonzero numbers. Prove that if limsup | an+1 / an | <1, then an converges absolutely. and prove that liminf | an+1 / an | > 1, then an diverges

Solution

Let A R and suppose that for each nN there is a function fn : AR. Then is called a sequence of functions on A. For each x A, this sequence gives rise to a sequence of real numbers, namely the sequence < fn(x) > . Definition. Let A R and let < fn > be a sequence of functions on A. Let A0 A and suppose f : A0 tends to R. Then the sequence is said to converge on A0 to f if for each xA0, the sequence converges to f(x) in R. In such a case f is called the limit function on A0 of the sequence . When such a function f exists, we say that the sequence is convergent on A0 or that converges pointwise on A0 to f and we write f(x) = n tends to infinity lim fn(x), xA0. Similarly, it summetion of fn(x) converges for every x A0, and if f(x) = infinity n=1 fn(x), x A0, the function f is called the sum of the series summetion of fn.

from the theorem :

Let {an} be a bounded sequence such that every convergent subsequence of {an} has a limit L. Prove that limn tends to infinity an = L. Solution. Method 1: Note that La = {L}. Hence lim supn an = lub(La) = L = glb(La) = lim infn tends to infinity an. So by Theorem 20.4, limn tends to infinity an = L. Method 2: Suppose for a contradiction that {an} does not have limit L. Hence there exists an > 0 such that for any integer N, there exists some n > N with |an L| > . This allows us to define n1 < n2 < n3 . . . such that |ani L| > for all i. Since {ani } is a bounded sequence, by Bolzano-Weierstrass it has a convergent subsequence, which clearly does not converge to L. This is a contradiction, and so it must be that limn tends to infinity an = L.