Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
14.2 Series
Any rational number can be written as a finite sum of fractions: 0.123 =
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 sum will be infinite: π = 3 +
1
4
1
5
9
2
+ 2 + 3 + 4 + 5 + 6 +++
10 10 10 10 10 10
∞
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
j=1
∞
⎪⎧ n ⎪⎫
partial sums as {Sn }n=1 = ⎨∑ a j ⎬ . If the sequence is convergent then S = lim Sn and the
n→∞
⎩⎪ j=1 ⎭⎪n=1
∞
series is called convergent. Otherwise the series is divergent.
∞
n
∑ k = lim ∑ k = lim
Ex 1.
n→∞
k=1
Ex 2.
∞
1
∑2
k=1
⎧
2
n→∞
k=1
Compute
n ( n − 1)
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 ⎫
,
+
=
=
,
+
+
=
=
,
+
+
+
=
=
,...,
⎨
⎬
4
4 2 4 8
8
8 2 4 8 16
16
16
2n ⎭
⎩2 2 4
∞
2n − 1
2x − 1
2 x ln 2
1
= lim x
= lim
= 1⇒ ∑ k = 1.
so we need to computer lim
n→∞ 2 n
n→∞ 2
L' Hospital x→∞ 2 x ln 2
k=1 2
Ex 3.
∞
∑
Compute
k=1
n
1
n2 + k
n
need to computer lim ∑
n→∞
k=1
n
Note, lim ∑
n→∞
k=1
1
n +k
2
n
≠ ∑ lim
k=1
n→∞
. The partial sums are given as Sn = ∑
k=1
1
n +k
2
1
n +k
2
.
= 0 since n isn’t finite.
1
n2 + k
, so we
Course: Accelerated Engineering Calculus I
1←
n
n
n +n
2
=∑
k=1
n
1
n +n
2
≤∑
k=1
Instructor: Michael Medvinsky
n
1
n +k
2
≤∑
k=1
1
n +1
2
=
n
n
→ 1 ⇒ lim ∑
n +1
2
∞
k=1
Geometric series: ∑ ar n = a + ar + ar 2 + ... =
Ex 4.
n=0
1
n +k
2
=1
a
when r < 1, otherwise is
1− r
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
∞
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→∞
n→∞
The converse theorem doesn’t true, see example below.
Ex 6.
A hyper harmonic series
∞
1
∑ n p converges for p > 1 , e.g.
n=1
1 π2
∑ n 2 = 6 . For
n=1
∞
∞
1
n=1 n
1
n
p=1, even though lim
= 0 , the harmonic series ∑ = ∞ is diverges since
n→∞
k
1 1
1 k 1 1 k
1
1 k ( k − 1) k − 1
≥ ⇒ Sk = ∑ ≥ ∑ = ∑1 = (1+ 2 ++ k ) =
=
n k
k n=1
k
k
2
2
n=1 n
n=1 k
k −1
S = lim Sk ≥ lim
=∞
k→∞
k→∞ 2
k≥n⇒
∞
The test for Divergence: If lim an ≠ 0 or DNE then S = ∑ an is divergent.
n→∞
n=1
∞
2n
2n
2x
2 x ln 2
lim
=
lim
=
lim
=∞
is
diverges
since
∑ 3n + 6
n→∞ 3n + 6
x→∞ 3x + 6
x→∞
3
n=1
Ex 7.
Thm: If
∞
∑ an and
n=1
constant) and
∞
∑(a
n
n=1
∞
∑ bn are convergent series, then so are the series
n=1
∞
∞
n=1
n=1
± bn ) . Furthermore: ∑ can = c∑ an and
∞
∑(a
n
n=1
∞
∑ ca
n
n=1
∞
∞
n=1
n=1
± bn ) = ∑ an ± ∑ bn .
(c is