1549901369-Elements_of_Real_Analysis__Denlinger_

58 Chapter 2 • Sequences

I

3n

2

Solution: (a) We want an n - 4n I

0 EN^3 n ~no=? n 2 + 5 - 3 < .01. Now,

I

3n

2

• 4n _ 31 = I (3n

2

• 4n) - 3 (n

2

+ 5) I

n^2 + 5 n^2 + 5

l

-4n - 151

n^2 + 5

4n + 15

n^2 + 5 ·

Thus, our objective is to find an no E N 3

4n+ 15

n ~ no =? n 2 +

5

< .01.

The latter inequality will be true if

n^2 + 5

---> 100 , i.e.,

4n+ 15

n^2 + 5 > 400n + 1500

n^2 - 400n > 1495

n(n - 400) > 1495.

Note that 404( 404 - 400) = 404 · 4 = 1616 > 1495. Thus,

I

qu11¢•'/tc. n ~ 404 =? n(n - 400) ~ 404(4) > 1495.

Thus, we take no = 404. The above analysis shows that

I

3n

2

• 4n I
  n ~ 404 =? n 2 + 5 - 3 < .01.

(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. Thus,

we want to find an no E N 3 n ~ no =? I 3n

2

4
n - 31 < .0005. As shown
n^2 + 5
above,

I

3n

2

• 4n _ 31 = 4n + 15.
  n^2 + 5 n^2 + 5

Thus, our objective is to find an no E N 3

4n+ 15

n ~ no =? n 2 +

5

<. 0005.