Formulas and Theorems 439
(2) If
∑∞
k= 1
akdiverges, then
∑∞
k= 1
bkdiverges.
(Note that if the smaller series diverges,
then the bigger series diverges.)
(d) Limit Comparison Test:
Given
∑∞
k= 1
akand
∑∞
k= 1
bkwith
ak>0,bk>0 for allks, and
letp=klim→∞
ak
bk
,if0<p<∞, then both
series converge or both series diverge.
(e) Integral Test:
Given
∑∞
k= 1
ak,ak>0 for allks, and
ak=f(k)for some function f(x),
if the function fis positive, continuous,
and decreasing for allx≥1, then
∑∞
k= 1
ak
and
∫∞
1
f(x)dx, either both converge or
both diverge.
- Maclaurin Series:
f(x)=
∑∞
k= 0
f(k)( 0 )
k!
xk
=f( 0 )+ f′( 0 )x+
f′′( 0 )
2!
x^2
+···+
f(k)( 0 )
k!
xk+···
sinx=
∑∞
k= 0
(− 1 )k
x^2 k+^1
( 2 k+ 1 )!
=x−
x^3
3!
+
x^5
5!
−
x^7
7!
+··· x∈R
cosx=
∑∞
k= 0
(− 1 )k
x^2 k
( 2 k)!
= 1 −
x^2
2!
+
x^4
4!
−
x^6
6!
+··· x∈R
ex=
∑∞
k= 0
xk
k!
= 1 +x+
x^2
2!
+
x^3
3!
+
x^4
4!
+··· x∈R
1
1 −x
=
∑∞
k= 0
xk
= 1 +x+x^2 +x^3 +··· x∈(−1, 1)
1
1 +x
=
∑∞
k= 0
(− 1 )kxk
= 1 −x+x^2 −x^3 +···+(−1)kxk+···
x∈(−1, 1)
ln( 1 +x)=
∑∞
k= 0
(− 1 )kx
k+ 1
k+ 1
=x−
x^2
2
+
x^3
3
−
x^4
4
+···
x∈(−1, 1]
tan−^1 x=
∑∞
k= 0
(− 1 )k
x^2 k+^1
2 k+ 1
=x−
x^3
3
+
x^5
5
−
x^7
7
+···
x∈[−1, 1]
- Taylor Series:
f(x)=
∑∞
k= 0
f(k)(a)
k!
(x−a)k
=f(a)+f′(a)(x−a)
+
f′′(a)
2!
(x−a)^2 +···
+
f(k)(a)
k!
(x−a)k+···
