1549901369-Elements_of_Real_Analysis__Denlinger_

Answers & Hints for Selected Exercises 643

n! 1 n

100

I Yn+l I (n + 1)

100

(c) Let Xn = n!

(^100) n and Yn = -Xn = -n. 1 -. Then --Yn = ( n + 1 )'. n 100

• ( n+- l)lOO · --^1 , 0. By Thm. 2.3.10, Yn _, 0. Apply Thm. 2.4.4.
  n n+l

9. (a) an=bn=n (b) an=2n, bn=n ( c) an = n, bn = 2n
   (d) an=n+L, bn=n


10. By Defn. 2.2. 12 , 3L > 0 and :3n1 E N 3 n 2: n1 ::::} bn = lbnl > L.
    M
    Let M > 0. Since an -> oo, :3n 2 E N 3 n 2: n 1 ::::} an > y· Then n >
    max{n1, n2}::::} anbn > M.
    1


11. Use t he given condit ion and Thm. 2.3.10 to show - -> 0. Apply Thm.
    Xn
    2.4.4.


12. Assume r > 1. For a > 0, a+ ar + ar^2 + · · · + arn > na -> +oo.
    For a < 0, a+ ar + ar^2 + · · · + arn = a(l + r + r^2 + · · · + rn) < na -> -oo.
    Apply Thm. 2.4.7.


13. Suppose an -> oo, bn -> L < 0, and let M > 0. Then 3 n1, n2 E N 3
    L 2M
    n 2: n1 ::::} bn < 2 and n 2: n2 ::::} an > _ L. Then
    n 2: max{n1, n2}::::} an(-bn) > M::::} anbn < -M.

EXERCISE SET 2 .5

2. In each of the following, Xn denotes the general term of the given sequence.

(a) Neither. X2n-l < X2n while X2n > X2n+1

... (-l)n+l (-l)n 1 1
(c) Strictly mcreasmg. Xn+l - xn = 1+ - --2: 1---- -
n+l n n+l n
2n + 1 2n + 2 2
= 1 - ( ) > 1 - ( ) = 1 - - > 0 when n > 2.
nn+l nn+l n
(e) Neither. {xn} = {1, 3, 1, 3, 1, 3, · · · , 1, 3, · · · }.

(h) Both monotone increasing and monotone decreasing: {O, 0, 0, 0, 0, 0, · · · }

Xn+i n + 1 2n ( n + 1) 1

(i) Strictly decreasing. Xn > 0 and ---;;:---=

2

n+l ·--;;:;:-= - n- · 2 < 1.

7r 7r

(k) Strictly increasing. As n _, oo, - decreases to 0, so cos -

2

increases
2n n
to 1.

(m) Strictly decreasing, since the function f(x) = 2

3

x +

5

has negative

x - x-2

derivative when x > 2.