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Unformatted text preview: Real Analysis II: Section 17 and 18 David Joseph Stith 17.6 Exercise. For each of the following prove or give a counterexample. (a) The statement, “If ( s n ) and ( t n ) are divergent sequences, then ( s n + t n ) diverges,” is false. Consider the sequences ( s n ) and ( t n ) defined by s n = − n and t n = n . Then ( s n ) and ( t n ) are divergent since ( s n ) and ( t n ) are unbounded but s n + t n = 0 for all n ∈ N so that ( s n + t n ) converges to zero. (b) The statement, “If ( s n ) and ( t n ) are divergent sequences, then ( s n t n ) converges,” is false. Consider the sequences ( s n ) and ( t n ) defined by s n = braceleftbigg − 1 , if n is odd; 1 , if n is even. and t n = s n for all n ∈ N . Then s n and t n both diverge, as has been previously shown, but s n t n = 1 for all n ∈ N so that ( s n t n ) converges to 1. (c) The statement, “If ( s n ) and ( s n + t n ) are convergent sequences then ( t n ) con verges,” is true. Suppose ( s n ) and ( t n ) are sequences such that ( s n ) converges and ( s n + t n ) converges. Let s = lim( s n ) and u = lim( s n + t n ). Now t n = ( s n + t n ) − ( s n ), so that by Theorem 17.1(a) and 17.1(b), lim(( s n + t n ) − ( s n )) = lim( s n + t n ) − lim( s n ) since both are both convergent = u − s Therefore ( t n ) converges, and in particular, lim( t n ) = lim( s n + t n ) − lim( s n ). Q.E.D. (d) The statement, “If ( s n ) and ( s n t n ) are convergent sequences, then ( t n ) con verges,” is false. Consider sequences ( s n ) and ( t n ) defined by s n = 0 and t n = n . Now ( s n ) converges, and s n t n = 0 for all n ∈ N so that ( s n t n ) converges, nevertheless ( t n ) is divergent since ( t n ) is unbounded. 17.7 Exercise. Give an example of an unbounded sequence that does not diverge to + ∞ or to −∞ . Consider the sequence ( s n ) defined by s n = ( − 1) n n . That ( s n ) is unbounded is evidenced by the fact that  s n  = n for all n ∈ N . Hence for any M ∈ R ,  s n  > M for all n > M . But ( s n ) does not diverge to + ∞ since for any M ∈ R , s n < M for all odd n > M , nor does ( s n ) diverge to −∞ since for any M ∈ R , s n > M for all even n > M . 1 Real Analysis II: Section 17 and 18 David Joseph Stith 17.8 Exercise. (a) Give an example of a convergent sequence ( s n ) of positive numbers such that lim( s n +1 /s n ) = 1 . Consider the sequence ( s n ) defined by s n = 1 n . Now s n is positive for all n ∈ N and ( s n ) is convergent (see Example 16.3). We will show that lim( s n +1 /s n ) = 1 by showing that for any ǫ > 0, there exists N ∈ R such that for n ∈ N , n > N = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle s n +1 s n − 1 vextendsingle vextendsingle vextendsingle vextendsingle < ǫ Suppose ǫ > 0. Then let N = 1 ǫ . Now, n > N = ⇒ n > 1 ǫ = ⇒ n + 1 > 1 ǫ = ⇒ 1 n + 1 < ǫ = ⇒ vextendsingle vextendsingle vextendsingle vextendsingle 1 n + 1 vextendsingle vextendsingle vextendsingle...
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 Spring '08
 Akhmedov,A
 Mathematical analysis, Tn, Dominated convergence theorem, David Joseph Stith

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