http://www.ck12.org Chapter 8. Infinite Series
Example 7
The sequence{ln(n)}grows without bound asnapproaches infinity. Note that the related functiony=ln(x)grows
without bound. The sequence is divergent because it does not have a finite limit. We write limn→+∞ln(n) = +∞.
Example 8
The sequence{ 4 −^8 n}converges to the limitL=4 and hence is convergent. If you graph the functiony= 4 −^8 nfor
n= 1 , 2 , 3 ,...,you will see that the graph approaches 4 asngets larger. Algebraically, asngoes to infinity, the term
−^8 ngets smaller and tends to 0 while 4 stays constant. We write limn→+∞( 4 −^8 n)=4.
Example 9
Does the sequencesnwith terms 1,− 1 , 1 ,− 1 , 1 ,− 1 ,....have a limit?
Solution
This sequence oscillates, or goes back and forth, between the values 1 and−1. The sequence does not get closer to
1 or−1 asngets larger. We say that the sequence does not have a limit, or limn→+∞sndoes not exist.
Note: Each sequence’s limit falls under only one of the four possible cases:
- A limit exists and the limit isL: limn→+∞sn=L.
- There is no limit: limn→+∞sndoes not exist.
- The limit grows without bound in the positive direction and is divergent: limn→+∞sn= +∞.
- The limit grows without bound in the negative direction and is divergent: limn→+∞sn=−∞.
If a sequence has a finite limit, then it only has one value for that limit.
Theorem
If a sequence is convergent, then its limit is unique.
Keep in mind that being divergent is not the same as not having a limit.
L’Hôpital’s Rule
Realistically, we cannot graph every sequence to determine if it has a finite limit and the value of that limit. Nor can
we make an algebraic argument for the limit for every possible sequence. Just as there are indeterminate forms when
finding limits of functions, there are indeterminate forms of sequences, such as^00 ,∞∞, 0 +∞. To find the limit of such
sequences, we can apply L’Hôpital’s rule.
Example 10
Find limn→+∞lnn(n).
Solution
We solved this limit by using a graph in Example 5. Let’s solve this problem using L’Hôpital’s rule. The numerator
is ln(n)and the denominator isn. Both functionsy=ln(n)andy=ndo not have limits. So, the sequence
{ln(n)
n}
is of the indeterminate form∞∞. Since the functionsy=ln(n)andy=nare not differentiable, we apply L’Hôpital’s
rule to the corresponding problem, limx→+∞lnx(x), first. Taking the first derivative of the numerator and denominator
ofy=lnx(x), we find limx→+∞lnx(x)=limx→+∞ 1 x^1 =0. Thus, limn→+∞lnn(n)=0 because the points ofy=lnn(n)are
a subset of the points of the functiony=lnx(x)asxapproaches infinity. We also confirmed the limit of the sequence
with its graph in Example 5.