Then the series converges if in VS .NET

Encoder QR-Code in VS .NET Then the series converges if

Then the series converges if
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If x 3 the series becomes
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1 1 X 1 1X1 , which diverges. 3n 3 n 1 n n 1 1 X 1 n 1 n 1
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If x 3 the series becomes
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1 1 X 1 n 1 , which converges. n 3 n 1
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Then the interval of convergence is 3 @ x < 3. The series diverges outisde this interval. Note that the series converges absolutely for 3 < x < 3. At x 3 the series converges conditionally. 1 n 1 x2n 1 (b) Proceed as in part (a) with un . Then 2n 1 !      n 2n 1 un 1  2n 1 !  2n 1 ! 2    lim  1 x x lim  lim  n!1 un  n!1 2n 1 ! 1 n 1 x2n 1  n!1 2n 1 ! lim 2n 1 ! x2 x2 lim 0 n!1 2n 1 2n 2n 1 ! n!1 2n 1 2n
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Then the series converges (absolutely) for all x, i.e., the interval of (absolute) convergence is 1 < x < 1.      n 1 ! x a n 1  u    un n! x a n ; lim  n 1  lim   lim n 1 jx aj: n!1 un  n!1  n!1 n! x a n This limit is in nite if x 6 a. Then the series converges only for x a. d un n x 1 n n 1 x 1 n 1 : n 3n 1 ; un 1 2 2n 1 3n 2 Then
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      u   n 1 3n 1 x 1  x 1 jx 1j    lim  n 1  lim    2  2 n!1 un  n!1 2n 3n 2 Thus, the series converges for jx 1j < 2 and diverges for jx 1j > 2. The test fails for jx 1j 2, i.e., x 1 2 or x 3 and x 1. 1 X n For x 3 the series becomes , which diverges since the nth term does not approach zero. 3n 1 n 1
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1 X 1 n n , which also diverges since the nth term does not For x 1 the series becomes 3n 1 n 1 approach zero. Then the series converges only for jx 1j < 2, i.e., 2 < x 1 < 2 or 1 < x < 3.
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INFINITE SERIES
287 b
11.26. For what values of x does (a)
a un   1 x 2 n : 2n 1 x 1
      u  2n 1 x 2 x 2     Then lim  n 1  lim x 1 if x 6 1; 2: n!1 un  n!1 2n 1 x 1
  1 x 2 n ; 2n 1 x 1 n 1
1 converge x n x n 1 n 1
      x 2 x 2 x 2      Then the series converges if  x 1 < 1, diverges if x 1 > 1, and the test fails if x 1 1, i.e., x 1. 2 If x 1 the series diverges. If x 2 the series converges. 1 X 1 n If x 1 the series is which converges. 2 2n 1  n 1  x 2  < 1, x 1 and x 2, i.e., for x @ 1. Thus, the series converges for    2 2   x 1 un 1  1  1, where un : However, noting that (b) The ratio test fails since lim  n!1 u  x n x n 1
1 1 1 x n x n 1 x n 1 x n we see that if x 6 0; 1; 2; . . . ; n,       1 1 1 1 1 1 Sn u1 u2 un x x 1 x 1 x 2 x n 1 x n 1 1 x x n and lim Sn 1=x, provided x 6 0; 1; 2; 3; . . . .
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