1549901369-Elements_of_Real_Analysis__Denlinger_

692 Appendix C • Answers & Hints for Selected Exercises

(b) "£bk converges by the alternating series test, but not absolutely, since

"£ lbkl = "£ ,h, a divergent p-series.

(c) By Mertens' Thm., their Cauchy product"£ Ck converges, but the con-

.. - I k (-l)k+l (-l)k+l I - k 1
vergence lS not absolute, smce ickl - i~ ~ (k+l-i)'/2 - i~ i3/2(k+l-i)1/2 2::

l3/2k1^1 / 2 -- v1k^1 > - k.^1

11. The kth term of the Cauchy product ("£ ak) ("£bk+"£ ck) is the kth
    k
    term of ("£ ak) ('L,(bk + ck)). By definition, this is "£ ai (bk+l-i + Ck+i-i) =
    i=l
    c~ aibk+l-i) + c~ aiCk+l-i), which is the sum of the kth term of("£ ak) ("£bk)
    and the kth term of("£ ak) ("£ck), so it must be the kth term of("£ ak) ("£bk)+
    ("£ ak) ("£ck)· Thus, the two series("£ ak) ("£bk+"£ ck) and("£ ak) ("£bk)+
    ("£ ak) ("£ck) have the same terms, so they must be equal.


12. (a) k~l c~ aibk+l-i) = a1b1 + (a1b2 + a2bi) + (a1b3 + a2b2 + a3b1) + ...

= a1 (b1 +b2+· · -+bn)+a2(b1 +b2+· · ·+bn-1)+a3(b1 +b2+· · ·+bn-2)+· · ·+anb1

n n

= a1Bn + a2Bn-l + · · · + anB1 = "£ akBn-k+l = "£ ak(B + Bn-k+I)·

k=l k=l

no n

(b) L lan+1-kl =Ian!+ lan-11 + lan- 21 + · · · + lan+l-nal = L lakl

k=l k=n+l-no

n oo

and L lan+i-kl < "£ la kl =A'.

k=no+l k=l

EXERCISE SET 8.5

3. See the solution to Exercise 8.3.12.


4. Let {ak}, {bk} be sequences. By Ex. 8.1.5, {ak} is the sequence of partial
   k
   sums of some 'L,xk. Then \fk EN, ak = "£xi. Applying Abel's summation by
   i=l n
   parts (8.5.2) with {xk} in place of {ak}, and with m = 1, we get "£ xkbk =
   k=l
   n n
   L ak(bk - bk+l) + anbn+l - (O)b1. That is, "£ bk(ak - ak+l) = anbn+l -
   k=l k=l
   n
   "£ ak(bk+i - bk)·
   k=l