Mathematics 120 Solutions for HWK 21b 2 function series that are geometric ∞ √x n 35. p671. For the series 1 , find the interval of convergence and, within this interval, 2 − n=0 find the sum of the seriesX as a function of x. Notice that this series isn’t actually a power series, but it is geometric. It has first term a = 1 and √x common ratio r = 1. It will converge absolutely provided r < 1, in other words provided 2 − | | √x that 1 < 1, and it will diverge otherwise. The convergence condition can be rewritten in | 2 − | √x the form 1 < 1 < 1. Solve for x as follows: − 2 − √x 1 < 1 < 1 − 2 − √x 0 < < 2 2 0 < √x < 4 0 < x < 16 Thus the interval of convergence is (0, 16). Using the formula for the sum of a geometric series, we can write the sum of the series as a function of x: ∞ √x n 1 2 1 = = 2 − √x 4 √x n=0 1 ( 1) − X − 2 − provided 0 < x < 16. 39. p671. For what values of x does the series 1 1 1 n 1 (x 3) + (x 3)2 + + (x 3)n + − 2 − 4 − ··· − 2 − ··· converge? What is its sum? What series do you get if you differentiate the given series term by term? For what values of x does the new series converge? What is its sum? 1 This series, which is a power series, is also geometric, with a = 1 and r = 2 (x 3). It converges 1 − − if and only if r < 1, in other words iff 2 (x 3) < 1. This last inequality can be rewritten in the form x |3| < 2, so the interval of convergence,| − − | centered at x = 3, is the interval 1 < x < 5. | − | a 1 2 The sum of the series is 1 r or 1 1 (x 3) , which can be simplified to give sum = x 1 . In other − 2 − words, we have −− − 1 1 1 n 2 1 (x 3) + (x 3)2 + + (x 3)n + = − 2 − 4 − ··· − 2 − ··· x 1 − Page 1 of 2 A. Sontag November 28, 1999 Math 120 HWK 21b solns continued provided 1 < x < 5. Differentiating gives 1 2 3 2 1 n n 1 2 + (x 3) (x 3) + + n − (x 3) − + = − −2 4 − − 8 − ··· 2 − ··· (x 1)2 − provided 1 < x < 5. Testing the endpoints x = 1 and x = 5 will easily show that the differentiated series still diverges at those points, so the interval of convergence for the differentiated series is also 1 < x < 5. Note. For both of these problems you could also find the interval of convergence by using the Ratio Test for Absolute Convergence and then testing endpoints. But there’s no need to, since these series are geometric. 40. p671. If you integrate the series in Exercise 39 term by term, what new series do you get? For what values of x does the new series converge, and what is another name for its sum? From Problem 39, we have 1 1 1 n 2 1 (x 3) + (x 3)2 + + (x 3)n + = − 2 − 4 − ··· − 2 − ··· x 1 − provided 1 < x < 5 Integrating gives 2 3 n+1 1 (x 3) 1 (x 3) 1 n (x 3) C + x − + − + + − + = 2 ln(x 1) − 2 2 4 3 ··· − 2 n + 1 ··· − provided 1 < x < 5. Setting x = 3 gives us C = 2 ln 2 3 = ln 4 3.So − − 2 3 n+1 1 (x 3) 1 (x 3) 1 n (x 3) x − + − + + − + = 2 ln(x 1) + 3 ln 4 − 2 2 4 3 ··· − 2 n + 1 ··· − − provided 1 < x < 5. There’s still a chance that the integrated series could converge at x = 1 or x = 5 or both. Testing x = 1 gives us a constant multiple of the harmonic series (write it out), hence a divergent series. Testing x = 5 gives an alternating harmonic series, hence a convergent series. So the interval of convergence is 1 < x 5. These methods don’t yet show, however, what ≤ the sum is when x = 5. Page 2 of 2 A. Sontag November 28, 1999.