1549901369-Elements_of_Real_Analysis__Denlinger_

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690 Appendix C • Answers & Hints for Selected Exercises


  1. L et ak = i.3.5.7 2·4·6 ·8 ....... .. (2k-1) ·(2k) · L et b k - l - ~ ak -^1 - 2k+1 2k+2 -_ 2k+2 i · s· mce """b u k
    diverges, ak ___, 0 by Ex. 48.


EXERCISE SET 8 .3


  1. This series has a grouping that diverges: (1 - 2) + ( ~ - 1) + a -~) +
    (~ -1) + · · · = -1 - ~ - ~ - ~ - · · · , so it diverges by Thm. 8.1.10.

  2. Prove the contrapositive: Suppose I: an converges.
    (a) If I: an converges absolutely, both I: a;t and I: a;;-converge by (a).
    (b) If I: an converges conditionally, both I: a;t and I: a;;- diverge by (b).

  3. By grouping pairs of successive terms we see that the even-numbered partial
    sums are S2n = E [(~)k - (~)k] = E (~)k - E (~)k, so S2n , 1-~ = ~·
    k=l k=l k=l
    For the odd-numbered partial sums, S2n+1 = S2n + 2 }+1
    , ~· Thus, by Ex.
    2.6.6, the series converges to ~·The series converges absolutely, since I: lanl =
    I: [(~)k + (~)k] =I: (~)k +I: (~)k.

  4. (a) Conv. but not abs. n > 9, 998. (b) Conv. but not abs. n > 31.
    (c) Diverges, by the general term test. (d) Converges absolutely. n ~ 5.
    (e) Conv. absolutely. n > 10. (f) Conv. but not abs. n > 646.
    2n n

  5. n~l + n~2 + · · · + 2 ~ = L f - L f = S2n from Ex. 7.
    k=l k=l

  6. (a),(d) converge conditionally (b),(e),(f) converge absolutely (c) diverges

  7. Suppose I: lakl converges and 't/n E N, lbkl :::; B. Then the partial sums
    n n
    Sn = L lakl are bounded above, say by B'. Then 't/n E N, L lakbkl <
    k=l k=l
    n oo
    BL iakl :::; BB'. .'. I: lakbkl converges, by Thm. 8.2.2.
    k=l k=l

  8. Consider {an}= {(1 + ~) - (~ + ~) + (i + ~) - (i + i) +···} .Then ,
    ( ) a an = 4 n24n 1 I 'f n IS · o dd , 4 4n-2 n2 4 n 'f I n · IS even.
    (b) an___, 0 (easy to show).


(c) For odd n ' an+1 an = 8n3+4n,8n (^3) +8n2 -2n-1 < l. For even n ' an+i an = 8n3+12n2-2n8n3-8n,-2n-1 - 3
< l. Thus, 't/n EN, an+i an < 1, so {an} is monotone decreasing.
(d) By (b), (c), and the alternating series test, I:ak converges.
(e) If Sn denotes the partial sum, then S4n =the sum of the first 4n terms
of the alternating harmonic series. Thus, by Ex. 7, S 4 n , ln 2, so Sn , ln 2.

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