Sequences Series And Convergence

2

Sequences, Series and Convergence:

2.1 Iterative Sequences 2.1.1

Exercises: Three iterative sequences are defined as follows:

(a)

a (1)  0, a (n  1)  (a(n)  5) / 8 ,

(c)

c(1)  1, c (n  1)  2 c(n)  1 ,

b (1)  0.5, b ( n  1)  8 b( n) /( b ( n) 2  1) ,

(b)

n  1, 2 , 3 ,

Complete the following table using your calculator. Record the answers only to 4 decimal places. (Hint: The following key sequences give the first row: 0 , = , ( Ans + 5 ) / 8 = , = , = , = , …) t(n+1) (a(n)+5)/8 8b(n)/(b(n)2+1) √(2c(n)+1)

2.1.2

t(1) 0.0000 0.5000 1.0000

t(2) 0.6250 3.2000 1.7321

t(3) 0.7031 2.2776 2.1128

t(4) 0.7129 2.9448 2.2860

t(5) 0.7141 2.4358 2.3605

t(10) 0.7143 2.6971 2.4136

t(15) 0.7143 2.6337 2.4142

t(25) 0.7143 2.6451 2.4142

Exercises: In the above exercise, continue the iteration, without recording, until two

consecutive iterates coincide. Then all the following iterates also will be the same. Obviously, this common iterate will be the limit of the sequence. Note down this limit to nine decimal places. Independently, by an analytic method, find the limit of the sequence. t(n+1) (a(n)+5)/8 8b(n)/(b(n)2+1)

Limit by Iteration 0.714285714 2.645751311

L=(L+5)/8 L = 8 L / (L2+1)

√(2c(n)+1)

2.414213562

L2 = 2 L + 1

2.1.3

Limit by Analytic Method L=5/7 0.714285714 √7 2.645751311 (1+√5 ) / 2

2.414213562

Exercises: By completing the table below, find the limit of the sequences approximately and

show that the convergence is linear. (a) a ( n  1)  (c)

1 , a(1)  1 a ( n) 2  1

c ( n  1) 

(1  c( n) 3 ) , c(1)  1 4

n

1

2

an

1.0000

an+1 - an

-0.5000

(an+2 - an+1)/( an+1 - an)

b(n  1)  sin(b(n))  1, b(1)  1

(b)

3

4

5

6

0.5000

0.8000

0.3000

-0.1902

7

8

0.6098

0.7290

0.1192

-0.0760

9

10

0.6530

0.7011

0.0481

-0.0306

0.6705

0.6899

0.6775

0.0194

-0.0123

0.0078

-0.6000 -0.6341 -0.6266 -0.6373 -0.6327 -0.6365 -0.6344 -0.6359 -0.6350 -0.6356

bn

1.0000

1.8415

1.9636

1.9238

1.9383

1.9332

1.9350

1.9344

1.9346

1.9345

bn+1 - bn

0.8415

0.1221

-0.0397

0.0145

-0.0051

0.0018

-0.0006

0.0002

-0.0001

0.0000

0.1451

-0.3255 -0.3643 -0.3525 -0.3569 -0.3554 -0.3559 -0.3557 -0.3558 -0.3558

1.0000

0.5000

(bn+2 - bn+1)/( bn+1 - bn) cn cn+1 - cn (cn+2 - cn+1)/( cn+1 - cn)

0.2542

0.2541

0.2541

0.2541

0.2541

0.2541

-0.5000 -0.2188 -0.0257 -0.0014 -0.0001

0.0000

0.0000

0.0000

0.0000

0.0000

0.4375

0.0484

0.0484

0.0484

0.0484

0.0484

0.1174

For the sequence (a) , the limit is

0.2813 0.0541

0.68

0.2556 0.0487

0.0484

correct to two decimal places. Also, (an+2 - an+1)/( an+1 - an)

is a constant (= - 0.6350) for sufficiently large n. Therefore, the convergence is linear.

2.1.4

Notes: Sequences whose convergence is linear converge slowly. However, there is a method

to accelerate the speed of convergence and to find the limit with high accuracy. Consider a sequence whose convergence is linear.

certain value N ,

a  a n 1    k ( say) . This implies that beyond a Suppose, lim  n 2 n   a n1  a n 

a n 2  a n1  k . We can therefore assume that a n 2  a n1  k a n1  a n  for all a n1  a n

n  N . It follows that if i is a positive integer, then a N  i 1  a N  i  k a N  i  a N  i 1   k 2 a N  i 1  a N  i  2     k i a N 1  a N   k i 1 a N  a N 1  .

Therefore, a N  i 1  (a N  i 1  a N  i )  a N  i  a N  i 1   a N  i 1  a N  i  2     a N 1  a N   a N  a N 1   a N 1





 a N  a N 1  k i 1  k i  k i 1    k  1  a N 1 .

Taking limits on both sides as i   , we get, lim a n = a N  a N 1  lim ( k i 1  k i    k  1)  a N 1 n

a N

=

 a N 1  lim (1  k    k i  k i 1 )  a N 1 = n 

n 

a N  k a N 1 a N  a N 1  a N 1 = 1 k 1 k

The above formula gives us a means of finding the limit fast and accurate. 2.1.5

Exercises: Using the above formula find the limits of the three sequences in Exercise 2.1.3 ,

correct to four decimal places. We complete the above table again with nine d.p. accuracy. n

6

7

8

9

10

an

0.652999725

0.701061373

0.670471796

0.689877632

0.677538381

an+1 - an

0.048061648

-0.030589577

0.019405836

-0.012339251

0.007835212

(an+2 - an+1)/( an+1 - an)

-0.636465421

-0.634393745

-0.635852590

-0.634982755

-0.635558539

bn

1.933218657

1.935040754

1.934393196

1.934623688

1.934541691

bn+1 - bn

0.001822098

-0.000647559

0.000230493

-0.000081997

0.000029176

(bn+2 - bn+1)/( bn+1 - bn)

-0.355391920

-0.355940832

-0.355745954

-0.355815345

-0.355790662

cn

0.254105133

0.254101855

0.254101696

0.254101689

0.254101688

cn+1 - cn

-0.000003278

-0.000000159

-0.000000008

0.000000000

0.000000000

(cn+2 - cn+1)/( cn+1 - cn)

0.048426439

0.048425784

0.048425756

0.048425710

0.048426433

We notice that for the sequence (a), we have

N  10 , k  0.635, a 9  0.689877632 and a10  0.677538381 . formula, we get

a N  k a N 1 1 k

lim a n 

n

=

0.67753831 0.636 (0.689877632 ) = 0.6823 1.635

Similarly, find the limits of the other two sequences:

lim bn 

n

,

lim c n 

n

Therefore, applying the

,