http://www.ck12.org Chapter 8. Infinite Series
n→lim+∞( 11
n −8
n^2)
=n→lim+∞^11 n −nlim→+∞n^82 Applying the rule for the difference of two limits=11 limn→+∞^1 n−8 limn→+∞n^12 Applying the rule for the limit of c times a limit
= 11 × 0 − 8 × 0 =0 Evaluating the limitsAs with limits of functions, there is a Sandwich/Squeeze Theorem for the limits of sequences.
Sandwich/Squeeze Theorem
Let{an},{bn}and{cn}be sequences. LetNbe a positive integer.
Supposecnis a sequence such thatan≤cn≤bnfor alln≥N. Suppose also that
liman=limbn=L. Then limcn=L.
You can see how the name of the theorem makes sense from the statement. After a certain point in the sequences,
the terms of a sequencecnare sandwiched or squeezed between the terms of two convergent sequences with the
same limit. Then the limit of the sequencecnis squeezed to become the same as the limit of the two convergent
sequences. Let’s look at an example.
Example 13
Prove limn→+∞^8 nn!=0.
Solution
Recall thatn! is read as “n factorial” and is written asn!=n×(n− 1 )×(n− 2 )×...× 1.
We want to apply the Sandwich theorem by squeezing the sequence^8 nn!between two sequences that converge to the
same limit.
First, we know that 0≤^8 nn!. Now we want to find a sequence whose terms greater than or equal to the terms of the
sequence^87 for somen.
We can write
8 n
n!=8 × 8 × 8 ...× 8
n×(n− 1 )×(n− 2 )×... 1
=^8 n×n−^81 ×...×^82 ×^81=( 8
n)( 8
n− 1 ×...×8
9 ×
8
8
)( 8
7 ×
8
6 ×
8
5 ×...×
8
1
)
Since each factor in the productn−^81 ×...× 98 ×^88 is less than or equal to 1, then the productn−^81 ×...×^89 ×^88 ≤1.
Then we make an inequality:
( 8
n)( 8
n− 1 ×...×8
9 ×
8
8
)( 8
7 ×
8
6 ×
8
5 ×...×
8
1
)
≤
( 8
7
)
( 1 )
( 8
7 ×
8
6 ×
8
5 ×...×
8
1
)
=
( 8
7
)( 8
7 ×
8
6 ×
8
5 ×...×
8
1
)
=
( 8
n