Given:
The sequence is
an=2+(−1)nn
. (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=2+(−1)11 =2−1 =1
Thus, the first term of the sequence is
a1=1.0000_
.
Substitute 1 for n in equation (2).
a2=2+(−1)22 =2+12 =52 =2.5
Thus, the second term of the sequence is
a2=2.5000_
.
Substitute 3 for n in equation (1).
a3=2+(−1)33=2−13=53=1.6667
Thus, the third term of the sequence is
a3=1.6667_
.
Substitute 4 for n in equation (1).
a4=2+(−1)44 =2+14 =94 =2.25
Thus, the fourth term of the sequence is
a4=2.2500_
.
Substitute 5 for n in equation (1).
a5= 2+(−1)55 =2−15 =95 =1.8
Thus, the fifth term of the sequence is
a5=1.8000_
.
Substitute 6 for n in equation (1).
a6=2+(−1)66 =2+16 =136 =2.1667
Thus, the sixth term of the sequence is
a6=2.1667_
.
Substitute 7 for n in equation (1).
a7=2+(−1)77 =2−17 =137 =1.8571
Thus, the seventh term of the sequence is
a7=1.8571_
.
Substitute 8 for n in equation (10.
a8=2+(−1)88 =2+18 =178 =2.125
Thus, the eighth term of the sequence is
a8=2.1250_
.
Substitute 9 for n in equation (1).
a9=2+(−1)99 =2−19 =179 =1.8889
Thus, the ninth term of the sequence is
a9=1.8889_
.
Substitute 10 for n in equation (1).
a10=2+(−1)1010 =2+110 =2110 =2.1
Thus, the tenth term of the sequence is
a5=2.1000_
.
Therefore, the first ten terms of the sequence are tabulated below:
n 
an=2+(−1)nn

1 
1.0000 
2 
2.5000 
3 
1.6667 
4 
2.2500 
5 
1.8000 
6 
2.1667 
7 
1.8571 
8 
2.1250 
9 
1.8889 
10 
2.1000 
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→∞(2+(−1)nn)=limn→∞(2)+limn→∞((−1)nn)
=2+limn→∞((−1)nn)
(3)
Compute the value of
(−1)nn
.
(−1)nn=(−1)nn ={1nif n is even−1nif n is odd =1n
The absolute value of
(−1)nn
when
n→∞
is obtained as follows:
limn→∞(−1)nn=limn→∞(1n)=1∞=0
Therefore,
limn→∞(−1)nn=0
Since
limn→∞(−1)nn=0
and by using Theorem (2), the value of
limn→∞((−1)nn)=0
.
Substitute
limn→∞((−1)nn)=0
in equation (3).
limn→∞an=2+0 =2
Therefore, the limit of the sequence is 2.