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# 115sols3 - HW 3 Solutions Math 115 Winter 2009 Prof Yitzhak...

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HW 3 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson 7.4 a) Let x n = 2 n . Then lim x n = 0 (as you may easily check - for ǫ > 0, just let N be ǫ 2 ), which is rational, even though all the x n are irrational. b) Consider the number 2 and its decimal expansion 1.41421..., and th3n let x 1 = 1 . 4, x 2 = 1 . 41, x 3 = 1 . 414, et cetera. The error | x n - 2 | is less than 10 - n at every step, and so if ǫ > 0, pick N = - log 10 ǫ ; then if n > N , | x n - 2 | < 10 log 10 ǫ = ǫ . Thus the x n converge to the irrational number 2 even though the x n themselves are all rational. 7.5 a) Find lim s n where s n = n 2 + 1 - n (no formal proof). To do this, un-rationalize the denominator by multiplying by n 2 +1+ n n 2 +1+ n ; we see that s n = n 2 +1 - n 2 n 2 +1+ n = 1 n 2 +1+ n . Now the denominator is greater than n 2 + n = 2 n , so we know that 0 < s n < 1 2 n . But lim 1 / 2 n = 0, and lim 0 = 0, so this means that lim s n = 0. b) Find lim s n where s n = n 2 + n - n (no formal proof). As with a), un-rationalize the denominator by multiplying and divid- ing by the conjugate n 2 + n + n . We see that s n = n n 2 + n + n . Divide both the numerator and denominator by n to get that s n = 1 1+ 1 n +1 . Now as n goes to infinity, 1 /n goes to zero, so the denominator gets closer and closer to 1 + 1 = 2, so then lim s n itself is 1 2 . c) Find lim s n where s n = 4 n 2 + n - 2 n (no formal proof). First un-rationalize the denominator, and then as in part b), divide both the numerator and the denominator by n . You should get that s n = 1 4+ 1 n +2 . As n gets bigger, the denominator goes to 4 + 2 = 4, so we see that lim s n = 1 4 .. 8.3. Let s n be a sequence of nonnegative real numbers, and suppose that lim s n = 0. Prove that lim s n = 0. To do this, verify the definition of the limit. Pick ǫ > 0. Then ǫ 2 > 0; since lim s n = 0, we may find N such that n > N implies that | s n | < ǫ 2 . s n is positive, so s n = | s n | < ǫ 2 . Take the square root of both sides to find that s n < ǫ , and thus | s n | < ǫ . So we have found N such that if n > N , then | s n | < ǫ ; therefore lim s n = 0. squaresolid 1

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2 8.4. Let ( t n ) be a bounded sequence; i.e. there exists M such that | t n | ≤ M for all n , and let ( s n ) be a sequence with lim s n = 0. Prove that lim( s n t n ) = 0. Pick ǫ > 0. Then as long as M > 0, ǫ/M > 0 as well (if M is zero, the t n are all zero, so the theorem is obvious. M cannot be negative because then no such t n exist). Since lim s n = 0, there exists an N such that n > N implies that | s n | < ǫ/M . But then if
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