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11.21. Investigate the convergence of
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(a) Here un n4 e n . Then   2    4 n 1 2  un 1  n 1 4 e n 2n 1    lim  n 1 e 2 lim   lim  2 n!1 un  n!1  n!1 n4 e n n4 e n  4  4 n 1 n 1 lim e 2n 1 lim lim e 2n 1 1 0 0 n!1 n!1 n!1 n n Since 0 < 1, the series converges. (b) Here un 1 n 1 2n . Then n2       1 n 2n 1 u  n2 2n2   lim lim  n 1  lim   u  n!1 n 1 2 1 n 1 2n  n!1 n 1 2 2 n!1  n Compare Problem 11.19(c).
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Since s > 1, the series diverges. (c) Here un 1 n . n2 1
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CHAP. 11]
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INFINITE SERIES     n 2 2   un 1    lim  1 n 1 n 1  lim n 1 n 1 1 lim   n!1 n 1 2 1 1 n 1 n n!1 n2 2n 2 n n!1 un  
and the ratio test fails. By using other tests [see Problem 11.19(a)], the series is seen to be convergent.
MISCELLANEOUS TESTS 11.22. Test for convergence 1 2r r2 2r3 r4 2r5 where (c) r 4=3.
(a) r 2=3,
(b) r 2=3,
  u  Here the ratio test is inapplicable, since  n 1  2jrj or 1 jrj depending on whether n is odd or even.  u  2 However, using the nth root test, we have p n jun j ( p p n 2jrn j n 2 jrj p n jrn j jrj if n is odd if n is even
Then lim
p n jun j jrj (since lim 21=n 1).
Thus, if jrj < 1 the series converges, and if jrj > 1 the series diverges. Hence, the series converges for cases (a) and (b), and diverges in case (c).
11.23. Test for convergence
    u  3n 1 2 The ratio test fails since lim  n 1  lim 1. n!1 un  n!1 3n 3
 2       1 1 4 2 1 4 7 2 1 4 7 . . . 3n 2 2 . 3 3 6 3 6 9 3 6 9 . . . 3n
However, by Raabe s test,
(      ) un 1  3n 1 2 4  lim >1 lim n 1   u  n!1 n 1 3n 3 n!1 3 n and so the series converges.
 2       1 1 3 2 1 3 5 2 1 3 5 . . . 2n 1 2 11.24. Test for convergence . 2 2 4 24t 2 4 6 . . . 2n
    u  2n 1 2 The ratio test fails since lim  n 1  lim  u  n!1 2n 2 1. n!1 n Also, Raabe s test fails since
(      ) u  2n 1 2 1 lim n 1  n 1  lim n 1  u  n!1 n!1 2n 2 n However, using long division,     un 1  2n 1 2 1 5 4=n 1 cn    u  2n 2 1 n 4n2 8n 4 1 n n2 where jcn j < P n so that the series diverges by Gauss test.
INFINITE SERIES
[CHAP. 11
SERIES OF FUNCTIONS 11.25. For what values of x do the following series converge a
1 X xn 1 ; n 3n n 1
1 X 1 n 1 x2n 1 n 1
2n 1 !
1 X n 1
n! x a n ;
1 X n x 1 n : 2n 3n 1 n 1
(a) un
xn 1 . n 3n
Assuming x 6 0 (if x 0 the series converges), we have     u   xn n 3n  n jxj n 1  lim lim  n 1  lim   n!1 3 n 1 jxj 3 n!1 un  n!1 n 1 3n 1 x jxj jxj < 1, and diverges if > 1. 3 3 jxj 1, i.e., x 3, the test fails. 3
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