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.)$a_{n}=\arctan \left(\frac{n^{2}}{n^{2}+4}\right)$

Problem 59E

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.)

$a_{n}=\arctan \left(\frac{n^{2}}{n^{2}+4}\right)$

Step-by-Step Solution

To determine

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

Answer

The sequence is convergent to the limit π4≈0.7854.

Explanation

Given:

The sequence is an=arctan(n2n2+4).

Definition used:

If an is a sequence and limn→∞an exists, then the sequence an is said to be convergent otherwise it is divergent.

Result used: Limit chain rule

If limu→bf(u)=L and limx→ag(x)=b with f(x) is continuous at x=b, then the value of limx→af(g(x)) is L.

Calculation:

Obtain the first 20 terms of the sequence.

n	n2n2+4	an=arctan(n2n2+4)
1	15=0.2000	arctan(0.2000)=0.2027
2	48=0.5000	arctan(0.5000)=0.4636
3	913=0.6923	arctan(0.6923)=0.6055
4	1620=0.8000	arctan(0.8000)=0.6747
5	2529=0.8621	arctan(0.8621)=0.7115
6	3640=0.9000	arctan(0.9000)=0.7328
7	4953=0.9245	arctan(0.9245)=0.7462
8	6468=0.9412	arctan(0.9412)=0.7551
9	8185=0.9529	arctan(0.9529)=0.7613
10	100104=0.9615	arctan(0.9615)=0.7658
11	121125=0.9680	arctan(0.9680)=0.7691
12	144148=0.9730	arctan(0.9730)=0.7717
13	169173=0.9768	arctan(0.9768)=0.7737
14	196200=0.9800	arctan(0.9800)=0.7753
15	225229=0.9825	arctan(0.9825)=0.7766
16	256260=0.9846	arctan(0.9846)=0.7776
17	289293=0.9863	arctan(0.9863)=0.7785
18	324328=0.9878	arctan(0.9878)=0.7793
19	361365=0.9890	arctan(0.9890)=0.7799
20	400404=0.9891	arctan(0.9901)=0.7799

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 59E

From the graph, it is observed that the sequence is convergent.

Also, the value of the limit can be guessed to lies between 0.7 and 0.8 as the plotted points are closer to between 0.7 and 0.8.

To prove:

The sequence an=arctan(n2n2+4) converges.

Proof:

Obtain the limit of the sequence (the value of the term an as n tends to infinity).

That is, compute limn→∞an=limn→∞(arctan(n2n2+4)).

Let g(n)=n2n2+4 and f(u)=arctan(u).

Obtain limn→∞g(n).

limn→∞g(n)=limn→∞(n2n2+4)

Divide the numerator and the denominator by the highest power.

limn→∞(n2n2+4)=limn→∞(n2n2n2+4n2)=limn→∞(1n2n2+4n2)limn→∞(11+4n2)

Redefine the terms as follows:

limn→∞(n2n2+4)=limn→∞(1)limn→∞(1)+limn→∞(4n2)=11+limn→∞(4n2)

Apply infinity property limn→∞(kna)=0 and simplify the terms.

limn→∞(n2n2+4)=11+1∞=11+0=11=1

Obtain limu→1f(u).

limu→1f(u)=limu→1(arctan(u))=arctan(1)=π4≈0.7854

Apply the limit chain rule and obtain the limit.

Since g(n)=n2n2+4 and f(u)=arctan(u), limn→∞(n2n2+4)=π4 and limn→∞(arctan(n2n2+4))=π4.

Therefore, limn→∞f(g(n))=π4.

Since limn→∞an exists and by using the definition, it can be concluded that the sequence is converging to the limit π4.

Therefore, the sequence is convergent to the limit π4≈0.7854.