Power Series

by: Engr. Ferdinand L. Marcos

Infinite Series

Sequence – succession of no. formed

according to some fixed rule.

Ex. 1, 8, 27, 256 is a sequence having the

rule that the nth is given by n3.

F. L. Marcos 2
 Infinite Series
Series – the indicated sum of a sequence
numbers.
Finite Series – series with numbers of terms
limited.
Infinite Series – series with numbers of
terms unlimited.

Note: Sometimes it is not practical to simplify

the series to identify the general term.
F. L. Marcos 3
Example 1. Find the first three terms and the
th n −1
(n + 1) term of the series whose nth term is 2n+1.
Solution:
𝑛−1
un = 2𝑛+1
1−1
u1 = =0
2 1 +1
2−1 1
u2= =
2 2 +1 5
3−1 2
u3 = =
2 3 +1 7
𝑛+1 −1 𝑛
un+1 = =
2 𝑛+1 +1 2𝑛+3
1 2 𝑛−1 𝑛
0+ 5 + 7 + . . . + 2𝑛+1 + 2𝑛+3 +...
F. L. Marcos 4
Example 2. Find the first four terms and the (n+1)th
2𝑛
term whose nth term is .
𝑛(𝑛+1)

F. L. Marcos 5
Example 3. Find the general term for the series:
1 1 1
1 + + + +. ..
4 9 16

F. L. Marcos 6
Example 4. Find the general term for the series:
𝑥 𝑥2 𝑥3
1+ + + +…
2! 3! 4!

F. L. Marcos 7
Example 5. Find the general term for the series:
3 5 7 9
- + -
1(2) 3(4) 5(6) 7(8)

F. L. Marcos 8
Convergent and Divergent series
Let the sequence s1, s2, ……sn denote the
respective partial sums of the series u1, u2, u3,
+ . . . . + un that is
S1= u1 s2 = u1 s3 = u1 + u2 + u3….
An infinite series is said to converge or
be convergent if the partial sums sn have a
definite limit s as n→∞; otherwise, the series
is said to diverge or to be divergent.

F. L. Marcos 9
Ratio test

Suppose that
𝑢𝑛+1
lim 𝑢 =k
𝑛→∞ 𝑛

Then:
If k<1, the series converges
If k>1, the series diverges
If k=1, the test fails

F. L. Marcos 10
Example 1. Determine the convergence or divergence of the
1 1 1
series 1 + 2! + 3! + . .. . + 𝑛! using ratio test.
Solution:
1
un =
𝑛!
1
un+1 = 𝑛+1 !
1
𝑛+1 !
lim 1
𝑛→∞ 𝑛+1 !
𝑛!
lim
𝑛→∞ 𝑛+1 !
1
lim
𝑛→∞ 𝑛+1
1
=∞ = 0
0<1 the series converges F. L. Marcos 11
Example 2. Determine the convergence or divergence of
2 2 2 2 3 2 4
the series - + (- ) + (- 3 ) + (- 3 ) using ratio test.
3 3

F. L. Marcos 12
Example 3. Determine the convergence or divergence of
2 3 4
the series + 32 +33 using ratio test.
3

F. L. Marcos 13
Root test

𝑛
lim 𝑢𝑛 = k
𝑛→∞
If k<1, the series converges
If k>1, the series diverges
If k=1, the test fails

F. L. Marcos 14
Example 1. Determine the convergence or
1 1 2 1 3 1 4
divergence of the series 2 + (2) + (2) + (2) using
root test.
Solution:
1 n
un (2)
1
1𝑛 𝑛 1
lim 2
=2 < 1; convergent series
𝑛→∞

F. L. Marcos 15
Example 2. Determine the convergence or
1 1 1
divergence of the series (ln 2) + (ln 3)2 + (ln 4 )3 +
1
using root test.
(ln 5)4

F. L. Marcos 16
Example 3. Determine the convergence or
−33 35 37
divergence of the series + - + . . . using root
12 24 36

test.

F. L. Marcos 17
Interval Convergence
An infinite series of the form a0, a1x +
a2x2 + . . . +anxn+ . . . in which a0, a1, a2, . . . an
are constant and x is a variable is called a
power series in x. The totality of values of x
for which a power series converges is called
its internal of convergence. This interval
always includes the value of x=0 and its range
can be determined by ratio test.

F. L. Marcos 18
Example 1. Find the interval of convergence of
the series: 1 + x + x2 + . . . . + xn-1

Solution:
un = xn-1
un+1 = xn
𝑥𝑛
lim 𝑥 𝑛−1
𝑛→∞
= 𝑥
The series will converge when -1<x<1
F. L. Marcos 19
Example 2. Find the interval of convergence of
𝑥2 𝑥3 𝑥4
the series: x + 2! + + 4! +..
3!

F. L. Marcos 20
Example 3. Find the interval of convergence of
the series: x + 2! x2 + 3! x3 + 4! x4 . . .

F. L. Marcos 21
Maclaurin’s Series
Assuming that a function f(x) can be
represented within a certain interval of
convergence by a power series of the form
f(x) = a0 + a1x + a2x2 + . . . . + anxn + . . . . → 1
where the a’s are constant to be determined.
Setting x=0 in 1, then f(0) = a0
Hence, the first coefficient a0 in 1 is
determined.

F. L. Marcos 22
Assuming that the series in 1 may be differentiated
term by term, and that this differentiation may be
continued, then:
f ‘ (x) = a1 + 2a2x + 3a3x2 + ……
f “ (x) = 2a2 + 6a3x + . . . . .
f ‘’’ (x) = 6a3 + . . . .

letting x=0 the results are

f’(0) = a1
f’’(0) = 2! a2
f ‘’’(0) = 3! a3
fn (0) = n! an
F. L. Marcos 23
solving for the constant
a0= f(0)
a1= f’(0)
a2= f”(0) / 2!
a3 = f ‘’’(0) / 3!

Substituting the value of a’s to 1

𝑓′′(0) 𝑓𝑛 (0)
f(x) = f(0) + f’(0)x + 𝑥2 +...+ +...
2! 𝑛!
This series called Maclaurin’s series or the power
series expansion of f(x) about x=0 it is named after
the Scottish mathematician.
F. L. Marcos 24
Example 1. Obtain the Maclaurin’s series expansion of ex

Solution:
f(x) = ex f(0) = 1
f ‘ (0) = ex f’(0) = 1
f ‘’ (0) = ex f ‘’ (0) =1
f ‘’’(0) = ex f ‘’’(0) =1
f ‘’’’(0) = ex f’’’’(0) = 1
1 𝑥2 1 𝑥3 1 𝑥4
ex = 1 + (1)x + + + +...
2! 3! 4!
𝑥2 𝑥3 𝑥4
ex = 1 + x+ + + +....
2! 3! 4!
or
∞ 𝑥 𝑛−1
ex = σ𝑛=1
𝑛−1 !

F. L. Marcos 25
Example 2. Compute e1/2 using Maclaurin’s
series.

F. L. Marcos 26
Example 3. Obtain the Maclaurin’s series
expansion of sin x.

F. L. Marcos 27
Example 4. Obtain the Maclaurin’s series
expansion of ln (1+x).

F. L. Marcos 28
Algebraic Operations in Power Series
The operations of algebra may be applied to power
series in the same manner that they are applied to
polynomials; provided that the series satisfy certain
condition of convergence.
Let a0 + a1x + a2x2 + a3x3+…+ anxn + …
b0 + b1x + b2x2 + b3x3+…+ bnxn + …
be convergent power series. The new convergent power
series may be attained as follows:
by adding or subtracting term by term
(a0 ± b0) + (a1 ± b1)x + . . . + (an ± bn)xn + . . .
by multiplication and grouping terms
(a0b0) + (a0b1 + a0b2)x + (a0b2 + a1b1+a2b1)x2 + . . .
F. L. Marcos 29
Example 1. Find the power series expansion of ½ (ex + e-x).
Solution:
𝑥2 𝑥3
ex = 1 + x + + 3! +....
2!
(−𝑥)2 (−𝑥)3 (−𝑥)4
ex = 1 – x + 2! + 3! + 4! +....
𝑥2 𝑥3 (−𝑥)2
½ (ex + e-x) = ½ [(1 + x + + 3! + . . .) + (1 – x + 2! +
2!
(−𝑥)3 (−𝑥)4
+ 4! +…)]
3!
2𝑥 2 2𝑥 4 2𝑥 6 2𝑥 8
½ (ex + e-x) = ½ [ 2+ 2! + 4! + 6! + 8!
𝑥2 𝑥4 𝑥6 𝑥8
½ (ex + e-x) = 1 + + + +
2! 4! 6! 8!
∞ 𝑥 2𝑛−2
σ𝑛=1
2𝑛−2
F. L. Marcos 30
Example 2. Find the power series expansion of
ex sin(x).

F. L. Marcos 31
Solution of Differential Equations using Power Series method
Steps:
1. Represent all given functions in the equation by power series in
powers of x. then, assume a solution in the form of a power series,
say
y= C0 + C1x + C2x2 + C3 x3 + . . .
2. Obtain term use differentiation
y’ = C1 + 2C2x + 3C3x2 + . . . .
y’’ = 2C2 + 6C3x + 4(3)C4x2 + . . .
3. Collect like powers of x
k0 + k1x + k2x2 + . . . = 0
where k0, k1 , k2 . . . are expressions containing the unknown
coefficients C0, C1 so that: k0 = 0 k1= 0 k2 = 0
4. Determine the coefficients C0, C1, C3, successively, then substitute
these coefficients into the assumed power series solution and
simplify. F. L. Marcos 32
Example 1. Solve y’ – y = 0 using power series method.
Solution:
Let y = C0 + C1x + C2x2 + C3x3 + C4x4 + . . .
y’= C1 + 2C2x + 3C3x2 + 4C4x3 + . . .
(C1 + 2C2x + 3C3x2 + 4C4x3 + ….) – (C0 + C1x + C2x2 + C3x3 + C4x4 + …)

C1 - C0 = 0 2C2 - C1 = 0 3C3 - C2 = 0
1 1
C1 = C0 C2 = ½ C1 = ½ C0 C3 = C2 = C0
3 6

4 C4 + C3 = 0
1 1
C4 = = C3 = C
4 24 0
1 1 1
y= C0 + C0x + C0x2 + C0x3 + C0x4 +....
2 6 24
𝑥2 𝑥3 𝑥4
y= C0 ( 1 + x + + + + ...)
2 6 24
𝑥2 𝑥3 𝑥4
y= C0 ( 1 + x + + + +...)
2! 3! 4!
y= C0 𝑒 𝑥 or y = C𝑒 𝑥 F. L. Marcos 33
Check using Differential Equations:
y’ – y = 0
dy
–y=0
dx
dy
=y
dx
dy
= dx
y
ln |y|= x + lnC
ln |y| - lnC = x
y
ln |c | = x
𝑦
ln | |
𝑒 𝑐 = 𝑒𝑥
𝑦 x
𝐶
= e
y = C ex F. L. Marcos 34
Example 2. Solve y’’ + y = 0 using power series method.

F. L. Marcos 35