Given:
The sequence is
an=sinnn
.
Definition used:
If
an
is a sequence and
limn→∞an
exists, then the sequence
an
is said to be convergent; otherwise it is divergent.
Theorem used: Squeeze Theorem
If
xn≤zn≤yn
for
n≥N
and
limn→∞xn=limn→∞yn=l
, then
limn→∞zn=l
. (1)
Calculation:
Obtain the first 20 terms of the sequence.
n 
sinn

an=sinnn

1 
sin1=0.8415

0.84151=0.8415

2 
sin2=0.9093

0.90932=0.4547

3 
sin3=0.1411

0.14113=0.0470

4 
sin4=−0.7568

−0.75684=−0.1892

5 
sin5=−0.9589

−0.95895=−0.1918

6 
sin6=−0.2794

−0.27946=−0.0466

7 
sin7=0.6570

0.65707=0.0939

8 
sin8=0.9894

0.98948=0.1237

9 
sin9=0.4121

0.41219=0.0458

10 
sin10=−0.5440

−0.544010=−0.0544

11 
sin11=−1.0000

−1.000011=0.0909

12 
sin12=−0.5366

−0.536612=−0.0447

13 
sin13=0.4202

0.420213=0.0323

14 
sin14=0.9906

0.990614=0.0708

15 
sin15=0.6503

0.650315=0.0434

16 
sin16=−0.2879

−0.287916=−0.0180

17 
sin17=−0.9614

−0.961417=−0.0566

18 
sin18=−0.7510

−0.751018=−0.0417

19 
sin19=0.1499

0.149919=0.0079

20 
sin20=0.9129

0.912920=0.0456

Plot the points
(n,an), for n=1,2,...20
on the graph as shown below in Figure 1.
From the graph, it is observed that the sequence is convergent.
Also, the value of the limit can be guessed to 0 as the plotted points are closer to zero.
To prove:
The sequence
an=sinnn
is convergent to 0.
Proof:
Obtain the limit of the sequence (the value of the term
an
as n tends to infinity).
That is, compute
limn→∞an=limn→∞sinnn
.
Since
−1≤sinn≤1
and divide by n,
−1n≤sinnn≤1n
Apply the Squeeze Theorem (1).
limn→∞(−1n)≤limn→∞(sinnn)≤limn→∞(1n)
Using infinity property
limn→∞(kna)=0
and obtain the limit.
0≤limn→∞(sinnn)≤0
Therefore,
limn→∞(sinnn)=0
Since
limn→∞an
is exist and by using the definition of the sequence, it can be concluded that the sequence is convergent to the limit 0.
Therefore, the sequence is convergent to the limit 0.
Hence the required proof is obtained.