364 STEP 4. Review the Knowledge You Need to Score High
∑∞
n= 1
1
√
n
, which is ap-series withp=
1
2
, and therefore diverges. Thus, the interval of
convergence is [−2, 0).
- Approximate the functionf(x)=
1
x+ 2
with a fourth degree Taylor polynomial
centered atx=3.
Answer: f(3)=
1
5
,f′(x)=
− 1
(x+2)^2
⇒f′(3)=
− 1
25
,
f′′(x)=
2
(x+2)^3
⇒f′′(3)=
2
125
,f′′′(x)=
− 6
(x+2)^4
⇒ f′′′(3)=
− 6
625
,
f(4)(x)=
24
(x+2)^5
⇒ f(4)(3)=
24
3125
,so
P(x)=
1 / 5
0!
(x−3)^0 +
− 1 / 25
1!
(x−3)^1 +
2 / 125
2!
(x−3)^2
+
− 6 / 625
3!
(x−3)^3 +
24 / 3125
4!
(x−3)^4
=
1
5
−
x− 3
25
+
(x−3)^2
125
−
(x−3)^3
625
+
(x−3)^4
3125
.
- Find the MacLaurin series for the functionf(x)=e−xand determine its interval of
convergence.
Answer:Sinceex=
∑xn
n!
, substitute−xto find
e−x=
∑(−x)n
n!
= 1 −x+
x^2
2
−
x^3
6
+···. The ratio limn→∞
∣∣
∣
∣
(−x)n+^1
(n+1)!
n!
(−x)n
∣∣
∣
∣
=nlim→∞
∣∣
∣∣ −x
n+ 1
∣∣
∣∣=0, so the series converges on the interval (−∞,∞).
14.9 Practice Problems
For problems 1–5, determine whether each
series converges or diverges.
1.
∑∞
n= 0
5 −n
2.
∑∞
n= 1
1
n· 2 n
3.
∑∞
n= 0
n
en
4.
∑∞
n= 1
n+ 1
n(n+2)
5.
∑∞
n= 1
n
(n+1)n