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x e" 0}, so since n-1 ’!0 as n ’! ", we have n-1/2 ’!0 as n ’! "; the result we proved
in Exercise 2.5.
2.21. Exercise. What do you deduce about the sequence an =exp (1/n) if you apply this
ex
result to the continuous function f(x) = ?
1
2.22. Example. Let an = for n e" 2. Show that an ’! 0 as n ’! ".
n log n
Solution. Note that 1 d" log n d" n if n e" 3, because log(e) = 1, log is monotone increasing,
and 2
1 1 1 1 1 1
n2 n log n n n n2 n log n
Here we have used the eventually form of the squeezing lemma.
1
2.23. Exercise. Let an = " for n e" 2. Show that an ’! 0 as n ’! ".
n log n
2.5. BOUNDED SEQUENCES 19
2.5 Bounded sequences
2.24. Definition. Say that {an} is a bounded sequence iff there is some K such that
|an| d"K for all n.
This definition is correct, but not particularly useful at present. However, it does provide
good practice in working with abstract formal definitions.
1
2.25. Example. Let an = " for n e" 2. Show that {an} is a bounded sequence. [This
n log n
is the sequence of Exercise 2.23].
2.26. Exercise. Show that the sum of two bounded sequences is bounded.
2.27. Proposition. An eventually bounded sequence is bounded
Proof. Let {an} be an eventually bounded sequence, so there is some N, and a bound K
such that |an| d"Kfor all n e" N. Let L =max{|a1|, |a2, . . . |aN-1|, K}. Then by definition
|a1| d" L, and indeed in the same way, |ak| d" L whenever k
|an| d"Kd"L, so in fact |an| d"Lfor all n, and the sequence is actually bounded.
2.28. Proposition. A convergent sequence is bounded
Proof. Let {an} be a convergent sequence, with limit l say. Then there is some N such
that |an - l|
, our pre-declared error, to be 1. Then by the triangle inequality,
|an| d"|an -l| +|l| d"1+|l| for all n e" N.
Thus the sequence {an} is eventually bounded, and so is bounded by Prop 2.27.
And here is another result on which to practice working from the definition. In order to
tackle simple proofs like this, you should start by writing down, using the definitions, the
information you are given. Then write out (in symbols) what you wish to prove. Finally
see how you can get from the known information to what you need. Remember that if a
definition contains a variable (like in the definition of convergence), then the definition is
true whatever value you give to it even if you use /2 (as we did in 2.10) or /K, for any
constant K. Now try:
2.29. Exercise. Let an ’! 0 as n ’! "and let {bn} be a bounded sequence. Show that
anbn ’! 0 as n ’! ". [If an ’! l =0 as n ’! ", and {bn} is a bounded sequence, then in
general {anbn} is not convergent. Give an example to show this.]
2.6 Infinite Limits
2.30. Definition. Say that an ’! "as n ’! "iff given K, there is some N such that
an e" K whenever n e" N.
This is just one definition; clearly you can have an ’!- "etc. We show how to use
such a definition to get some results rather like those in 2.10. For example, we show
n2 +5n
an = ’! " as n ’! ".
3n+2
20 CHAPTER 2. SEQUENCES
n +5
Pick some K. We have an = n. =n.bn (say). Using results from 2.10, we see
3n+2
that bn ’! 1/3 as n ’! ", and so, eventually, bn > 1/6 (Just take = 1/6 to see this).
Then for large enough n, we have an >n/6 >K, providing in addition n >6K. Hence
an ’! " as n ’! ".
Note: It is false to argue that an = n.(1/3) ’! "; you can t let one part of a limit
converge without letting the other part converge! And we refuse to do arithmetic with "
so can t hope for a theorem directly like 2.10.
Chapter 3
Monotone Convergence
3.1 Three Hard Examples
So far we have thought of convergence in terms of knowing the limit. It is very helpful to
be able to deduce convergence even when the limit is itself difficult to find, or when we can
only find the limit provided it exist. There are better techniques than we have seen so far,
dealing with more difficult examples. Consider the following examples:
n
1
e
Sequence definition of : Let an = 1+ ; then an ’!e as n ’! ".
n
"
n!
Stirling s Formula: Let an = " ; then an ’! 2À as n ’! ".
n(n/e)n
Fibonacci Sequence: Define p1 = p2 =1, andfor ne"1, let pn+2 = pn+1 + pn.
"
pn+1 1+ 5
Let an = ; then an ’! as n ’! ".
pn 2
We already saw in 2.12 that knowing a sequence has a limit can help to find the limit.
As another example of this, consider the third sequence above. We have
pn+2 pn 1
=1 + and so an+1 =1 + . (3.1)
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