1549901369-Elements_of_Real_Analysis__Denlinger_

470 Chapter 8 • Infinite Series of Real Numbers

Proof. Let {an} be a sequence of positive real numbers.

• 
  
  
  
  • an+l
    Part 1: We prove first that lim y1a;; s; lim --.
    n--+cx:> n--+(X) an
    Let L = lim an+l. If L = +oo, there is nothing to prove. So, we suppose
    n--+oo an
    L < +oo. Choose any M > L. Then by the c-criterion for upper limits (see
  
  
  
  

Theorem 2.9 .7) 3n 0 EN 3 n::;:: n 0 => an+l < M. Thus, Vk EN,
an

That is,

ano+k < M ano+(k-l) < M2ano+(k-2) < ... < Mkano.

n >no=> a < Mn-noa =Mn __ ano
n no Mno
n f;:, ano
=> y1a;; < My"' where c = Mno.
Now, lim ye= 1 (see Example 2.3.9). Thus, since upper limits preserve
n-><Xl
inequalities,
lim y1a;; s; M lim ye = M.
n--+ cx::> n--+ oo
Hence, by the forcing principle (1.5.9), lim y1a;; s; L.
n-><Xl

Part 2: The proof that lim an+i s; lim y1a;; is left as Exercise 40. •

n-+oo an n--+oo

Theorem 8.2.18 shows that the root test is at least as strong as the ratio
test, in the sense that if the ratio test proves that a nonnegative series converges
(or diverges), the root test will also. The next example shows that the root test
is actually "stronger" than the ratio test. That is , there are nonnegative series
for which the ratio test is inconclusive but the root test works.

<Xl 1 1 1 1 1 1
Example 8.2.19 Let n=O 2:::: a n = 1+1+-+-+-+-+-+-+···. 2 3 22 32 23 33 That

. 1. 1
is, Vk E N, a2k =
2
k, while a2k+1 =
3
k.

When we attempt to use the ratio test on this series, we get

an+l

{

-b/ 2

1

k = (~)k if n = 2k, }

~ = 2 k~1 /-b = (~)k · ~ if n = 2k + 1.

Thus, the subsequence consisting of even-numbered terms of { a::i} converges

to 0, while the subsequence consisting of odd-numbered terms diverges to +oo.
So for the ratio test, L. = 0 and L = +oo, and the ratio test is inconclusive.

We now test this series using the root test.

If n = 2k, yra;; = 2~ = (2-k) i /2k = 2-i/ 2 = ~.