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
- A series
∑∞
n= 1
an is said toconverge absolutelyif the series of absolute values
∑∞
n= 1
|an|
converges.
- An alternating series
∑∞
n= 1
an is said toconverge conditionally if the series
∑∞
n= 1
an
converges but not absolutely.
- If a series converges absolutely, then it converges, i.e., if
∑∞
n= 1
|an|converges, then
∑∞
n= 1
an
converges.
- 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.