homework_6 - } -P | < 1 sup {| s n | : n...

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Homework 6 11.10. (a) Suppose s n k is a convergent subsequence of s n = (1 , 1 2 , 1 , 1 , 1 2 , 1 3 ,... ) . Let s = lim k →∞ s n k . Either s = 0 or s > 0, in which case we claim that s must have the form 1 n for n N . If s is not of the form 1 n for some n N then we must have strict inequalities 1 n + 1 < s < 1 n for some n . Our idea is to define δ = inf {| s - 1 n | : n N } to get a lower bound on the distances from s to potential terms of the subsequence s n k . Since s is closest to 1 n and 1 n +1 among all terms of the form 1 n , we have δ = inf {| s - 1 n | : n N } = min {| s - 1 n | , | s - 1 n + 1 |} > 0 . But then | s n k - s | ≥ δ for all k because each term of s n k has the form 1 n for n N , and this contradicts that lim k →∞ s n k = s . Hence, S = { 1 n : n N } ∪ { 0 } . (b) By theorem, we have lim sup s n = sup S = 1 and lim inf s n = inf S = 0 by (a). 12.10. There are two statements to prove. If s n is bounded by | s n | ≤ M for all n N where M is a real number, then we have lim sup | s n | ≤ sup {| s n | : n N } ≤ M < . On the other hand, if lim sup | s n | = P < then choosing ± = 1 we obtain an N such that | sup {| s n | : n > N
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Unformatted text preview: } -P | &lt; 1 sup {| s n | : n &gt; N } -P &lt; 1 sup {| s n | : n &gt; N } &lt; P + 1 . Hence, we can bound {| s n | : n N } by M = max {| s 1 | , | s 2 | ,..., | s N | ,P + 1 } . 12.14. (a) By Theorem 12.2 we have lim sup | n ! | 1 /n lim inf | n ! | 1 /n lim inf | ( n + 1)! n ! | = lim inf | n +1 | = so by the lim inf/lim sup theorem, we have that ( n !) 1 /n diverges to . 1 2 (b) Let s n = 1 n n ( n !). Then, | s n +1 s n | = n n ( n +1) n . for all n . We claim that lim n n n ( n +1) n = 1 e by the quotient theorem for limits because lim n n ( n + 1) n = lim 1 ( n +1) n n n = lim 1 (1 + 1 n ) n = 1 e . By Corollary 12.3, we then have lim 1 n ( n !) 1 /n = lim | 1 n n ( n !) | 1 /n = lim | s n +1 s n | = e-1 ....
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homework_6 - } -P | &amp;amp;lt; 1 sup {| s n | : n...

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