Then the series converges for all x except x 0; 1; 2; 3; . . . ; and its sum is 1=x. in .NET

Generate QR Code in .NET Then the series converges for all x except x 0; 1; 2; 3; . . . ; and its sum is 1=x.

Then the series converges for all x except x 0; 1; 2; 3; . . . ; and its sum is 1=x.
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UNIFORM CONVERGENCE 11.27. Find the domain of convergence of 1 x x 1 x x2 1 x .
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Method 1: Sum of first n terms Sn x 1 x x 1 x x2 1 x xn 1 1 x 1 x x x2 x2 xn 1 xn 1 xn If jxj < 1, lim Sn x lim 1 xn 1.
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n!1 n!1 n!1
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If jxj > 1, lim Sn x does not exist. If x 1; Sn x 0 and lim Sn x 0.
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If x 1; Sn x 1 1 n and lim Sn x does not exist.
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Thus, the series converges for jxj < 1 and x 1, i.e., for 1 < x @ 1. Method 2, using the ratio test. The series converges if x 1.   u  If x 6 1 and un xn 1 1 x , then lim  n 1  lim jxj. n!1 un  n!1 If x 1, the series
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Thus, the series converges if jxj < 1, diverges if jxj > 1. The test fails if jxj 1. converges; if x 1, the series diverges. Then the series converges for 1 < x @ 1:
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11.28. Investigate the uniform convergence of the series of Problem 11.27 in the interval (a) 1 < x < 1, (b) 1 @ x @ 1, c :99 @ x @ :99; d 1 < x < 1, 2 2 2 2 e 0 @ x < 2.
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[CHAP. 11
(a) By Problem 11.27, Sn x 1 xn ; S x lim Sn x 1 if 1 < x < 1; thus, the series converges in this 2 2 n!1 interval. We have Remainder after n terms Rn x S x Sn x 1 1 xn xn The series is uniformly convergent in the interval if given any  > 0 we can nd N dependent on , but not on x, such that jRn x j <  for all n > N. Now jRn x j jxn j jxjn <  when n ln jxj < ln  or n > ln  ln jxj
since division by ln jxj (which is negative since jxj < 1) reverses the sense of the inequality. 2 ln  ln  But if jxj < 1 ; ln jxj < ln 1 , and n > > 1 N. Thus, since N is independent of x, the 2 2 ln jxj ln 2 series is uniformly convergent in the interval. ln  ln  A N, so that the series is also uniformly (b) In this case jxj @ 1 ; ln jxj @ ln 1 ; and n > 2 2 ln jxj ln 1 1 1 2 convergent in @ x @ :
Reasoning similar to the above, with 1 replaced by .99, shows that the series is uniformly convergent in 2 :99 @ x @ :99. ln  (d) The arguments used above break down in this case, since can be made larger than any positive ln jxj number by choosing jxj su ciently close to 1. Thus, no N exists and it follows that the series is not uniformly convergent in 1 < x < 1. (c) (e) Since the series does not even converge at all points in this interval, it cannot converge uniformly in the interval.
11.29. Discuss the continuity of the sum function S x lim Sn x of Problem 11.27 for the interval n!1 0 @ x @ 1.
If 0 @ x < 1; S x lim Sn x lim 1 xn 1.
n!1 n!1
If x 1; Sn x 0 and S x 0. & 1 if 0 @ x < 1 Thus, S x and S x is discontinuous at x 1 but continuous at all other points in 0 if x 1 0 @ x < 1. In Problem 11.34 it is shown that if a series is uniformly convergent in an interval, the sum function S x must be continuous in the interval. It follows that if the sum function is not continuous in an interval, the series cannot be uniformly convergent. This fact is often used to demonstrate the nonuniform convergence of a series (or sequence).
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