5 Steps to a 5 AP Calculus BC 2019

Series 355

|S−s 3 |=| 3. 2 − 3. 25 |=|− 0. 05 |= 0 .05, which is clearly less thana 4 =

1

16

= 0 .0625. The

coefficient ofa 4 is negative, as isS−s 3.

Example 2

IfSis the sum of the series

∑∞

n= 1

(− 1 )n

n!

andsn, itsnthpartial sum, find the maximum value

of

∣

∣S−s 4

∣

∣.

Note that

∑∞

n= 1

(− 1 )n

n!

=−

1

1!

+

1

2!

−

1

3!

+...is an alternating series such thata 1 >a 2 >a 3 ...,

i.e.,an > an+ 1 and limn→∞an=nlim→∞

1

n!

=0. Therefore,|S−sn|≤an+ 1 , and in this case

|S−s 4 |≤a 5 , anda 5 =

1

5!

=

1

120

. Thus,|S−s 4 |≤

1

120

.

Example 3

The series

∑∞

n= 1

(−1)n

√

n

satisfies the hypotheses of the alternating series test, i.e.,an≥an+ 1 and

nlim→∞an=0. IfSis the sum of the series

∑∞

n= 1

(−1)n

√

n

andsnis thenthpartial sum, find the

minimum value ofnfor which the alternating series error bound guarantees that|S−sn|<
0 .01.

Since the series
∑∞
n= 1

(−1)n

√

n

satisfies the hypotheses of the alternating series test,|S−sn|≤an+ 1 ,

and in this case,an+ 1 =

1

√

n+ 1

. Setan+ 1 ≤ 0 .01, and you have

1

√

n+ 1

≤ 0 .01, which

yields 10, 000≤n+1. Therefore,n≥9999. The minimum value ofnis 9999.

Absolute and Conditional Convergence

1. A series

∑∞

n= 1

an is said toconverge absolutelyif the series of absolute values

∑∞

n= 1

|an|

converges.

2. An alternating series

∑∞

n= 1

an is said toconverge conditionally if the series

∑∞

n= 1

an

converges but not absolutely.

3. If a series converges absolutely, then it converges, i.e., if
   ∑∞
   n= 1

|an|converges, then

∑∞

n= 1

an

converges.

4. If

∑∞

n= 1

|an|diverges, then

∑∞

n= 1

anmay or may not converge.

Example 1

Determine whether the series− 1 +

1

3

−

1

9

+

1

27

···=

∑∞

n= 1

(−1)n

1

3 n−^1

converges.