Given:
The sequence is
an=1+(−12)n
. (1)
Theorem used:
If
limn→∞an=0
, then
limn→∞an=0
. (2)
Calculation:
Obtain the first ten terms of the sequence.
Substitute 1 for n in equation (1).
a1=1+(−12)1 =1−12 =12 =0.5
Thus, the first term of the sequence is
a1=0.5000_
.
Substitute 2 for n in equation (1).
a2=1+(−12)2 =1+14 =54 =1.25
Thus, the second term of the sequence is
a2=1.2500_
.
Substitute 3 for n in equation (1).
a3=1+(−12)3 =1−18 =78 =0.875
Thus, the third term of the sequence is
a3=0.8750_
.
Substitute 4 for n in equation (1).
a4=1+(−12)4 =1+116 =1716 =1.0625
Thus, the fourth term of the sequence is
a4=1.0625_
.
Substitute 5 for n in equation (1).
a5=1+(−12)5 =1−132 =3132 =0.96875
Thus, the fifth term of the sequence is
a5=0.9687_
.
Substitute 6 for n in equation (1).
a6=1+(−12)6 =1+164 =6564 =1.015625
Thus, the sixth term of the sequence is
a6=1.0156_
.
Substitute 7 for n in equation (1).
a7=1+(−12)7 =1−1128 =127128 =0.9921875
Thus, the seventh term of the sequence is
a7=0.9922_
.
Substitute 8 for n in equation (1).
a8=1+(−12)8 =1+1256 =257256 =1.00390625
Thus, the eighth term of the sequence is
a8=1.0039_
.
Substitute 9 for n in equation (1).
a9=1+(−12)9 =1−1512 =511512 =0.998046875
Thus, the ninth term of the sequence is
a9=0.9980_
.
Substitute 10 for n in equation (1).
a10=1+(−12)10 =1+11024 =10251024 =1.0009765625
Thus, the tenth term of the sequence is
a10=1.0010_
.
Therefore, the first ten terms of the sequence are tabulated below:
n 
an=1+(−12)n

1 
0.5000 
2 
1.2500 
3 
0.8750 
4 
1.0625 
5 
0.9688 
6 
1.0156 
7 
0.9922 
8 
1.0039 
9 
0.9980 
10 
1.0010 
Plot the points
(n,an), for n=1,2,...10
on the graph as shown below in Figure 1.
Obtain the limit of the sequence (the value of the term
an
as n tends to infinity).
limn→∞an=limn→∞(1+(−12)n) =limn→∞(1)+limn→∞(−12)n
=1+limn→∞(−12)n
(3)
Compute the value of
(−12)n
.
(−12)n=(−1)n2n ={12n,if n is even−12n,if n is odd =12n =(12)n
Apply
limn→∞
on both sides and simplify the terms.
limn→∞(−12)n=limn→∞(12)n =limn→∞enln(12)
=elimn→∞nln(12)
(4)
Obtain
limn→∞nln(12)
.
limn→∞nln(12)=ln(12)⋅limn→∞n =ln(12)⋅∞ =−ln2⋅∞
=−∞
(5)
Substitute equation (5) in equation (4) and obtain the value of
limn→∞(−12)n
.
limn→∞(−12)n=e−∞ =0
Since
limn→∞(−12)n=
0 and by using equation (2), the value of
limn→∞(−12)n
is 0.
Substitute
limn→∞(−12)n=0
in equation (3).
limn→∞an=1+0 =1
Therefore, the limit of the sequence is
1_
.