Analysis17and18

Analysis17and18 - Real Analysis II: Section 17 and 18 David...

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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 > 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 < Suppose > 0. Then let N = 1 . Now, n > N = n > 1 = n + 1 > 1 = 1 n + 1 < = vextendsingle vextendsingle vextendsingle vextendsingle 1 n + 1 vextendsingle vextendsingle vextendsingle...
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Analysis17and18 - Real Analysis II: Section 17 and 18 David...

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