1873_solutions

At n n the sequence x sin 1 n n decreases to zero for

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Unformatted text preview:  exp t log 1 and so all we have to show is that if we define x t for t x then the function g is increasing. Since g x  0 and since g is continuous on the interval 0 whenever t  x. Now x, Ý it will follow that g is increasing when we have shown that g t whenever t  x we have g t  t log 1 g t  log 1 x t  and therefore what we want to show is that the inequality log 1 u  u 1u holds whenever 0 u  1. We define h u  log 1 u  x t 1 x t 0 u 1u and observe that h 0  0 and that whenever 0  u  1 we have u 1  1 u u  hu 1u 1 u2 u1 Therefore h u  0 whenever 0  u  1 and our proof is complete. 2 b. Use the dominated convergence theorem to show that n Ý xn 1 n x  1  Þ e x x  1 dx  lim Þ n Ý 1/n 0 0 . For the definition of the gamma function , see an earlier exercise. Some Exercises on Power Series 1. Find the radius of convergence and the interval of convergence of each of the following series: a. n! 2 x n 2n ! An easy application of d’Alembert’s test shows that the series converges when |x |  4. To see what happens when x  4 we define n! 2 4 n an  2n ! for each n. Since n1 ! 2 4 n1 2 n 2 ! 2n  2n  2  1 2n  1 n! 4 2n ! we see that a n does not approach 0 as n Ý and so the given series must diverge when x  4. Thus the interval of convergence is 4, 4 and the radius of convergence is 4. a n 1  an b. 2n ! x n An easy application of d’Alembert’s test shows that the series converges when n! 2 |x |  1 . To see what happens when x  1 we define 4 4 2n ! an  n 4 n! 2 for each n. Since 372 2 n 2 ! 4 n1 n1 ! 2  1 2n  1  2n  1  1 2 n1 2n  2 2n ! n n! 2 4 and so the sequence a n is decreasing. Since n lim n 1 aa 1  1  1 nÝ n 2 Raabe’s test guarantees that a n is divergent but the criterion for terms to approach zero guarantees that a n 0 as n Ý. It therefore follows from Dirichlet’s testDirichlet’s test that, 1 n a n is convergent. The interval of convergence of although a n is divergent, the series n 2n ! x 11 is therefore , . the series 44 n! 2 a n 1  an c. n! x n nn An easy application of d’Alembert’s test shows that the series converges when |x |  e. To see what happens when x  e we define n! e n an  nn for each n. Since n  1 ! e n 1 n  1 n 1 n lim n 1 aa 1  n Ý n 1 lim nÝ n n! e n nn e 1 1 n  n Ýn 1 lim n  1 2 we know that the sequence a n does not approach zero and a n diverges. When x  e we have n! x n n! e n  1n n n nn and, once again, because this nth term fails to approach zero the series diverges. The interval n! x n of convergence of the series is therefore e, e . nn nnxn d. n! We saw in an earlier group of exercises that the interval of convergence of this series is 11 e, e . 2. Given that c 1 and that 1 1x 1, expand the function f in a power series with center c and find the interval of convergence of fx  whenever x this series. Hint: Use the identity 1 x 1 x 1   1 1 c 1 xc 1c . Thus 1 1 The series xc 1c n Ý 1 1 converges if and only if 373 c n0 x 1 c c n x 1...
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