S3 - 92.305 Homework 2 Solutions October 2, 2009 Exercise...

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Unformatted text preview: 92.305 Homework 2 Solutions October 2, 2009 Exercise 2.2.1. Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit. (a) lim 1 (6 n 2 +1) = 0. Proof: Let > 0 be given. We take any N N strictly greater than radicalbig ( 1 1) / 6. Then n N implies n 2 > ( 1 1) / 6, hence 6 n 2 > 1 1, and 6 n 2 + 1 > 1 . Therefore, 1 / (6 n 2 + 1) < . Noting also that 1 / (6 n 2 + 1) > > , we see that N works. squaresolid Proof: To keep the algebra simple, take N = radicalbig ( 1 ) / 6. (Note that 1 6 n 2 +1 < 1 6 n 2 , so if can make 1 6 n 2 less than , 1 6 n 2 +1 will be less than as well.) (b) lim (3 n +1) (2 n +5) = 3 2 . Proof: Let > 0 be given. We take any N N strictly greater than (13 10 ) / 4 . Then n N implies n > (13 10 ) / 4 , hence 4 n > 13 10 , and | 4 n + 10 | = (4 n + 10) = 4 n + 10 > 13 = | 13 | . Therefore, vextendsingle vextendsingle vextendsingle vextendsingle 3 n + 1 2 n + 5 3 2 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 6 n + 2 6 n 15 4 n + 10 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 13 4 n + 10 vextendsingle vextendsingle vextendsingle vextendsingle < , as desired. squaresolid (c) lim 2 n +3 = 0. Proof: Let > 0 be given. We take any N strictly greater than (2 / ) 2 3. Then n N implies n > (2 / ) 2 3, hence n + 3 > (2 / ) 2 and n + 3 > 2 / . This gives 2 / n + 3 < , and noting also that 2 / n + 3 > > shows that N works. squaresolid Exercise 2.2.4. Argue that the sequence 1 , , 1 , , , 1 , , , , 1 , , , , , 1 , 5 zeros , 1 , . . . does not converge to zero. For what values of > 0 does there exist a response N ? For which values of > 0 is there no suitable response?...
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S3 - 92.305 Homework 2 Solutions October 2, 2009 Exercise...

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