1549901369-Elements_of_Real_Analysis__Denlinger_

54 Chapter 2 • Sequences

1

Solution: (a) We want an no EN 3 n 2'. no::::} 2n+^3 21

3

n _

7

• 3 < .01. Now,

I

2n + 3 _ ~I = I 3(2n + 3) - 2 (3n - 7) I

3n - 7 3 3(3n - 7)

I

6n + 9 - 6n + 141

3(3n - 7)

23

l9n - 211 ·

We can eliminate the absolute value bars in the denominator if 9n- 21 > 0.

This inequality is true if 9n > 21, which is true when n 2'. 3. Thus, when n 2'. 3,

l9n - 211=9n-21. Thus, our objective now is to find an no EN 3 both no 2'. 3

and

23

n 2'. no ::::} 9n - 21 < .01.

The latter inequality will be true if

9n - 21

23 > 100

9n - 21 > 2300

9n > 2321

n > 257 .88 · · ·.

Take no = 258. We have shown that n > 258 ::::} 1

2

n +

3

• ~I < .01.


• 3n- 7 3
  (b) To make sure that the nth term of the sequence approximates the limit
  accurately to three decimal places, we want to guarantee that rounding off to
  three decimal places does not cause a change in the third decimal digit. That
  is , we want to guarantee that the nth term is within .0005 of the limit. So, we

want to find an no E N3 n 2'. no ::::} 1

2

n +

3

• ~I < .0005. As shown above, if
  3n- 7 3
  n 2'. 3,

I

2n + 3 21 23

3n - 7 - 3 = 9n - 21 ·

Thus, our objective is to find an n 0 E N 3 no 2'. 3 and

23 1

n 2'. no ::::} 9n - 21 <. 0005 = (^2) , 000.