Use a graph of the sequence to decide whether thesequence is convergent or divergent. If the sequence is convergent,guess the value of the limit from the graph and then proveyour guess. (See the margin note on page 739 for advice ongraphing sequences.)$a

Problem 58E

Use a graph of the sequence to decide whether thesequence is convergent or divergent. If the sequence is convergent,guess the value of the limit from the graph and then proveyour guess. (See the margin note on page 739 for advice ongraphing sequences.)

$a_{n}=\frac{\sin n}{n}$

Step-by-Step Solution

To determine

Whether the sequence is convergent or divergent; guess the value of the limit if convergent with valid proof.

Answer

The sequence is convergent to the limit 0.

Explanation

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.

Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals, 8th (James Stewart Calculus), Chapter 11.1, Problem 58E

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.