1549901369-Elements_of_Real_Analysis__Denlinger_

8.2 Nonnegative Series 467

Solution.

an+l. (n+l)^2 +1 2n. n^2 +2n+2 2n

(a) lim = hm · --= hm · --

n-+oo an n-+oo 2n+l n2 + 1 n-+oo n2 + 1 2n+l

= lim (

1

+ ~ + ~ · ~) = ~.

n-+oo 1 + ~ n 2 2

00 2+1

In this case, L = ~ < 1, so by the ratio test, L n

2

n converges.

n=l

(b) lim an+l = lim (n + 1)!. 3n = lim n + 1 = +oo.

n-+oo an n-+oo 3n+l n! n-+oo 3.

oo I

L

n.

In this case, L = +oo, so by the ratio test, -3n diverges.

n=l

(c) 1 Im. an+l --= 1 Im. n +^1 · ---

n-+oo an n-+oo (n + 1)3 + 1

n^3 +1

n

= lim n + 1. n

3

• 1 = 1.
  n-+oo n n^3 + 3n^2 + 3n + 2
  In this case, L = 1, so the ratio test gives no insight into the convergence
  oo n n 1.
  or divergence of'""' - 3 --. However, - 3 --< 2 so the companson test tells
  n=l ~n+l n+l n
  us that this series converges. D

The limit form of the ratio test cannot be used in cases where lim an+i
n-+OO an
fails to exist. This shortcoming can be overcome by using upper and lower
limits, defined in Section 2.9. The upper and lower limits always exist (but
may be infinite).^1

Theorem 8.2.13 (Ratio Test, Upper and Lower Limit Form) Suppose

6 """"' an is · a nonnega t ive · series · an d l t e _ L = l " Im --an+^1 an d -L = -1. Im --an+^1 ( pos-
n-+oo an n-+oo an
sibly +oo). Then

(a) ifL < 1, the series l:an converges, and

(b) if L. > 1, the series Lan diverges.

Proof. Exercise 37. •

1. The upper limit Jim Xk and lower limit Jim Xk of a sequence {xk} are defined in Section
   k-oo k-+oo
   2.9, where it is also proved that, whenever {xk} converges, Jim Xk = Jim Xk = Jim Xk-
   k-+oo k-oo k-+oo