Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 1.2 Series (8.2)
1
2
3
+ 2 + 3 . 10 10 10
A real number can also be written as sum of fractions, but when the number is irrational the 1
4
1
5
9
2
sum will be infinite: π = 3 + + 2 + 3 + 4 + 5 + 6 + + + 10 10 10 10 10 10
Any rational number can be written as a finite sum of fractions: 0.123 =
∞
Def: A sum of infinite sequence {an }n=1 , S = ∑ a j is called (infinite) series. ∞
j=1
∞
Def: In order to find a sum of infinite sequence S = ∑ a j one defines a sequence of partial sums j=1
∞
⎧⎪ n ⎫⎪
∞
as {Sn }n=1 = ⎨∑ a j ⎬ . If the sequence is convergent then S = lim Sn and the series is called n→∞
⎪⎩ j=1 ⎪⎭n=1
convergent. Otherwise the series is divergent. ∞
n
n ( n − 1)
k
=
lim
=∞
Ex 1.
∑ n→∞ ∑ k = lim
n→∞
2
k=1
k=1
Ex 2.
Compute
∞
k=1
⎧
1
∑2
k
. The partial sums are given as
⎫
n
{S } = ⎨∑ 21 ⎬ =
n
⎩ k=1
k
⎭
⎧ 1 1 1 2 + 1 3 1 1 1 4 + 2 + 1 7 1 1 1 1 8 + 4 + 2 + 1 15
2n − 1 ⎫
= , + + =
= , + + + =
= ,..., n ⎬
⎨ , + =
4
4 2 4 8
8
8 2 4 8 16
16
16
2 ⎭
⎩2 2 4
∞
2n − 1
2x − 1
2 x ln 2
1
S n = lim n = lim x
= lim x
= 1⇒ S = ∑ k = 1.
thus we compute lim
n→∞
n→∞ 2
n→∞ 2
L' Hospital x→∞ 2 ln 2
k=1 2
Ex 3.
n
1
Find lim ∑
n→∞
n2 + k
k=1
1←
Ex 4.
n
n
=∑
n
1
1
. Note, that lim
n→∞
1
≤∑
n
≤∑
n2 + k
1
= 0 so it is convergent.
n
n
→ 1 ⇒ lim ∑
1
= 1 k=1
n2 + 1
n2 + k
∞
a
Geometric series: ∑ ar n = a + ar + ar 2 + ... =
when r < 1 , otherwise is
1− r
n=0
n2 + n
k=1
n2 + n
n2 + k
k=1
k=1
n2 + 1
=
divergent.
Ex 5.
Telescopic series
∞
1
∞
⎛1
1 ⎞
⎛
1⎞ ⎛ 1
1⎞
⎛ 1
1⎞ ⎛ 1
1 ⎞
1
∑ n ( n + 1) = ∑ ⎜⎝ n − n + 1⎟⎠ = ⎜⎝ 1− 2 ⎟⎠ + ⎜⎝ 2 − 3 ⎟⎠ + ...⎜⎝ n − 1 − n ⎟⎠ + ⎜⎝ n − n + 1⎟⎠ = 1− n + 1
n=1
n=1
4 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky ∞
Thm: If S = ∑ an is convergent series, then n=1
lim an = lim ( Sn+1 − Sn ) = lim Sn+1 − lim Sn = S − S = 0 n→∞
n→∞
n→∞
x n→∞
The converse theorem doesn’t true, see example below. Ex 6.
We will see later that a hyper harmonic series (p-series)
∞
1
∑n
p
is
n=1
1 π2
1
convergent for p > 1 , e.g. ∑ 2 =
. For p=1, even though lim
= 0 , the
n→∞
6
n
n=1 n
∞
∞
k
1
n=1 n
1
n=1 n
harmonic series ∑ = ∞ is diverges since it’s partial sum’s sequence Sk = ∑ has
a divergent subsequence S2 :
k
1
2
1 ⎛ 1 1⎞
1 ⎛ 1 1⎞
1 1
2
S4 = 1+ + ⎜ + ⎟ > 1+ + ⎜ + ⎟ > 1+ + = 1+
2 ⎝ 3 4⎠
2 ⎝ 4 4⎠
2 2
2
S2 = 1+
S8 = 1+
1 ⎛ 1 1⎞ ⎛ 1 1 1 1⎞
1 ⎛ 1 1⎞ ⎛ 1 1 1 1⎞
3
+ ⎜ + ⎟ + ⎜ + + + ⎟ > 1+ + ⎜ + ⎟ + ⎜ + + + ⎟ = 1+
2 ⎝ 3 4⎠ ⎝ 5 6 7 8⎠
2 ⎝ 4 4⎠ ⎝ 8 8 8 8⎠
2
1 ⎛ 1 1⎞ ⎛ 1 1 1 1⎞ ⎛ 1 1 1 1 1 1 1 1 ⎞
S16 = 1+ + ⎜ + ⎟ + ⎜ + + + ⎟ + ⎜ + + + + + + + ⎟ >
2 ⎝ 3 4 ⎠ ⎝ 5 6 7 8 ⎠ ⎝ 9 10 11 12 13 14 15 16 ⎠
1 ⎛ 1 1⎞ ⎛ 1 1 1 1⎞ ⎛ 1 1 1 1 1 1 1 1 ⎞
4
+ ⎜ + ⎟ + ⎜ + + + ⎟ + ⎜ + + + + + + + ⎟ = 1+
⎝
⎠
⎝
⎠
⎝
⎠
2
4 4
8 8 8 8
16 16 16 16 16 16 16 16
2
n
S2n > 1+ → ∞
2
> 1+
∞
The test for Divergence: If lim an ≠ 0 or DNE then S = ∑ an is divergent. n→∞
Ex 7.
∞
2
n
∑ 3n + 6
n=1
∞
∞
n=1
n=1
n=1
is diverges since lim
n→∞
n
2
2x
2 x ln 2
= lim
= lim
=∞
3n + 6 x→∞ 3x + 6 L ' Hospital x→∞ 3
∞
Thm: If ∑ an and ∑ bn are convergent series, then so are the series ∑ can (c is constant) and ∞
∑(a
n
n=1
n=1
∞
∞
∞
∞
∞
n=1
n=1
n=1
n=1
n=1
± bn ) . Furthermore: ∑ can = c∑ an and ∑ ( an ± bn ) = ∑ an ± ∑ bn . Ex 8.
Ex 9.
∞
∞
3 7
1
1
π2
π2
∑ n 2 + 2 n = 3∑ n 2 + 7∑ 2 n = 3⋅ 6 + 7 ⋅1 = 2 + 7
n=1
n=1
n=1
∞
∞
∑
n=0
1
( 2)
n
= 1+
∞
1 1
1
1
1
1
1
1
1 ∞ 1
+ +
+ 2+ 2
+ 3+ 3
+ ... = ∑ n +
∑
2 2 2 2 2 2 2 2 2 2
2 n=1 2 n
n=0 2
5