Analysis19and20-1

# Analysis19and20-1 - Real Analysis II Section 19 and 20...

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Unformatted text preview: Real Analysis II: Section 19 and 20 David Joseph Stith 19.10 Exercise. Suppose that x > 1 . Prove that lim x 1 n = 1 . Let s n = x 1 n . We will show that ( s n ) is bounded below and decreasing and hence convergent. Assuming that x 1 n represents the positive n th real root of x , we already know that x 1 n ≥ 0. To see that x 1 n > 1 suppose to the contrary that x 1 n ≤ 1. Then, x 1 n ≤ 1 = ⇒ ( x 1 n ) n ≤ 1 n since x 1 n ≥ = ⇒ x ≤ 1 which contradicts x > 1. Therefore it must be that x 1 n > 1 and hence ( s n ) is bounded below by 1. To show that ( s n ) is decreasing we will show that s n +1 − s n < 0. We have, s n +1 − s n = x 1 n +1 − x 1 n = x 1 n +1 parenleftBig 1 − x 1 n ( n +1) parenrightBig = s n +1 ( 1 − s n ( n +1) ) < 0 since s k > 1 for all k. Therefore ( s n ) is decreasing. Therefore, by the Monotone Convergence Theorem, ( s n ) is convergent. Now let s = lim( s n ). It remains only to show that s = 1. By Theorem 19.4 every subsequence of ( s n ) converges to s , therefore lim( s 2 n ) = s in particular, with s 2 n = x 1 2 n = radicalbig x 1 n = √ s n Therefore by Example 17.6, lim s 2 n = lim √ s n = √ lim s n = √ s and hence s = √ s . Therefore s must be either 0 or 1. But s n > 1 for all n , therefore it cannot be that s = 0. Therefore s = 1. Therefore lim x 1 n = 1. Q.E.D. 19.13 Exercise. Let ( s n ) and ( t n ) be bounded sequences. (a) Prove that limsup( s n + t n ) ≤ limsup( s n ) + limsup( t n ) . Let s = limsup( s n ) and let t = limsup( t n ). Then, by Theorem 19.11a, for any ǫ > 0, there exists N 1 such that n > N 1 = ⇒ s n < s + ǫ/ 2 and there exists N 2 such that n > N 2 = ⇒ t n < t + ǫ/ 2. Therefore, n > max { N 1 , N 2 } = ⇒ s n < s + ǫ 2 and t n < t + ǫ 2 = ⇒ s n + t n < s + t + ǫ 1 Real Analysis II: Section 19 and 20 David Joseph Stith We will use this result to show that limsup( s n + t n ) ≤ s + t ....
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Analysis19and20-1 - Real Analysis II Section 19 and 20...

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