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COMPARISON TEST AND QUOTIENT TEST 11.5. If 0 @ un @ vn ; n 1; 2; 3; . . . and if vn converges, prove that un also converges (i.e., establish the comparison test for convergence).
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Let Sn u1 u2 un ; Tn v1 v2 vn . Since vn converges, lim Tn exists and equals T, say.
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Also, since vn A 0; Tn @ T.
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Then Sn u1 u2 un @ v1 v2 vn @ T or 0 @ Sn @ T: Thus Sn is a bounded monotonic increasing sequence and must have a limit (see 2), i.e., un converges.
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11.6. Using the comparison test prove that 1 1 1 2 3
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We have
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1 1 4 5 1 1 9 10 1 2 1 6
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1 X1 n 1
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diverges.
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1A 1 3 1 7 1 15 A A A
1 2 1 1 1 4 4 2 1 1 1 1 1 8 8 8 8 2 1 1 1 16 16 16
1 16 (8 terms) 1 2
etc.
Thus, to any desired number of terms, 1 1 1 1 1 1 1 A 2 3 4 5 6 7
1 1 2 2
Since the right-hand side can be made larger than any positive number by choosing enough terms, the given series diverges. 1 X 1 By methods analogous to that used here, we can show that , where p is a constant, diverges if np n 1 p @ 1 and converges if p > 1. This can also be shown in other ways [see Problem 11.13(a)].
1 X n 1
11.7. Test for convergence or divergence
Since ln n < n and
ln n . 2n3 1
1 1 ln n n 1 @ 3 ; we have @ 3 2: 2n3 1 n 2n3 1 n n
1 X 1 converges. n2 n 1
Then the given series converges, since
11.8. Let un and vn be positive.
  u  By hypothesis, given  > 0 we can choose an integer N such that  n A <  for all n > N. Then for v  n n N 1; N 2; . . .  < un A< vn
1 X N 1
un constant A 6 0, prove that un converges or diverges vn according as vn converges or diverges. If lim
A  vn < un < A  vn
Summing from N 1 to 1 (more precisely from N 1 to M and then letting M ! 1), A  vn @
1 X N 1
un @ A 
1 X N 1
There is no loss in generality in assuming A  > 0. Then from the right-hand inequality of (2), un converges when vn does. From the left-hand inequality of (2), un diverges when vn does. For the cases A 0 or A 1, see Problem 11.66.
INFINITE SERIES
1 X 4n2 n 3 ; n3 2n n 1 1 X n p n ; 2n3 1 n 1 1 X ln n . n2 3 n 1
[CHAP. 11
11.9. Test for convergence;
4n2 n 3 4n2 4 4n2 n 3 4 (a) For large n, is approximately 3 . Taking un and vn , we have 3 n n n n 2n n3 2n u lim n 1. n!1 vn Since vn 4 1=n diverges, un also diverges by Problem 11.8. Note that the purpose of considering the behavior of un for large n is to obtain an appropriate comparison series vn . In the above we could just as well have taken vn 1=n. ! 4n2 n 3 4. Then by Theorem 1, Page 267, the series converges. Another method: lim n n!1 n3 2n p n n n 1 (b) For large n, un 3 is approximately vn 3 2 . 2n 2n 2n 1 X u 1X 1 vn converges ( p series with p 2), the given series converges. Since lim n 1 and n!1 vn n2 p 2  n n 1 . Then by Theorem 1, Page 267, the series converges. Another method: lim n2 n!1 2n3 1  2   ln n ln n ln n @ lim n3=2 (c) lim n3=2 2 lim p 0 (by L Hospital s rule or otherwise). Then by n!1 n!1 n!1 n n 3 n2 Theorem 1 with p 3=2, the series converges. ln n n 1 Note that the method of Problem 11.6(a) yields 2 < , but nothing can be deduced since n 3 n2 n 1=n diverges.
1 X n 1
11.10. Examine for convergence:
e n ;
1 X n 1
sin3
  1 . n
lim n2 e n 0 (by L Hospital s rule or otherwise). n!1 verges.
Then by Theorem 1 with p 2, the series con-
(b) For large n, sin 1=n is approximately 1=n. This leads to consideration of   & ' 1 sin 1=n 3 lim n3 sin3 lim 1 n!1 n!1 n 1=n from which we deduce, by Theorem 1 with p 3, that the given series converges.
INTEGRAL TEST 11.11. Establish the integral test (see Page 267).
We perform the proof taking N 1. Modi cations are easily made if N > 1. From the monotonicity of f x , we have un 1 f n 1 @ f x @ f n un n 1; 2; 3; . . .
Integrating from x n to x n 1, using Property 7, Page 92, n 1 f x dx @ un n 1; 2; 3 . . . un 1 @
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