Download Power Series Convergence: Guide with Examples and Problems and more Lecture notes Signals and Systems in PDF only on Docsity!
Joe Foster
Interval of Convergence of Power Series
Power Series : A power series about x = a is a series of the form
∞ ∑
n =
cn ( x − a )
n = c 0 + c 1 ( x − a ) + c 2 ( x − a )
2
n
in which the centre a and the coefficients c 0 , c 1 , c 2 ,... , cn,... are constants.
The Radius of Convergence of a Power Series : The convergence of the series
cn ( x − a )
n is described by one of
the following three cases:
- There is a positive number R such that the series diverges for x with | x − a | > R but converges absolutely for x with
| x − a | < R. The series may or may not converge at either of the endpoints x = a − R and x = a + R.
- The series converges absolutely for every x ( R = ∞)
- The series converges only at x = a and diverges elsewhere ( R = 0)
The Interval of Convergence of a Power Series : The interval of convergence for a power series is the largest interval
I such that for any value of x in I , the power series converges.
The interval of convergence can be calculated once you know the radius of convergence. First you solve the inequality
| x − a | < R for x and then you check each endpoint individually. So how do we calculate the radius of convergence? We
use the ratio test (or root test) and solve.
Example 1 - Geometric Power Series : Taking all the coefficients to be 1 in the power series centred at x = 0 gives
the geometric power series:
∞ ∑
n =
x
n = 1 + x + x
2
3
n
This is the geometric series with first term 1 and ratio x.
Sn = 1 + x + x
2
3
4
n
=⇒ (1 − x ) Sn = (1 − x )
1 + x + x
2
3
4
n
1 + x + x
2
3
4
n
x + x
2
3
4
5 · · · + x
n +
= 1 − x
n +
=⇒ Sn =
1 − x n
1 − x
So,
∞ ∑
n =
x
n = lim n →∞
Sn = lim n →∞
1 − x
n
1 − x
which converges if and only if | x | < 1
MATH 142 - Interval of Convergence of Power Series Joe Foster
Example 2 : Consider the power series
( x − 2) +
( x − 2)
2 − · · · +
) n
( x − 2)
n
Centre: a = 2, c 0 = 1, c 1 = −
, c 2 =
,... , cn =
) n
Ratio: r =
cn +1( x − 2) n +
cn ( x − 2) n
c 1 ( x − 2)
c 0
1 2 ( x − 2)
x − 2
The series converges when | r | < 1, that is,
x − 2
x − 2
< 1 =⇒ | x − 2 | < 2 =⇒ − 2 < x − 2 < 2 =⇒ 0 < x < 4_._
Example 3 : For what values of x do the following series converge?
(a)
∞ ∑
n =
n − 1 x
n
n
We will use the Ratio Test:
an +
an
n x
n +
n + 1
n
(−1) n −^1 xn
nx
x + 1
= | x |
n
n + 1
n →∞ −→ | x |
The series converges absolutely when | x | < 1 and diverges when | x | > 1. It remains to see what happens at the
endpoints; x = −1 and x = 1.
x = −1:
∞ ∑
n =
n − 1 (−1)
n
n
∞ ∑
n =
2 n − 1
n
∞ ∑
n =
n
∞ ∑
n =
n
=⇒ the series diverges at x = −1.
x = 1:
∞ ∑
n =
n − 1 1
n
n
∞ ∑
n =
n − 1
n
= the Alternating Harmonic Series =⇒ the series converges at x = 1.
So, the series
∞ ∑
n =
n − 1 x
n
n
converges for − 1 < x ≤ 1 and diverges elsewhere.
(b)
∞ ∑
n =
x
n
n!
We will use the Ratio Test:
an +
an
x
n +
( n + 1)!
n!
x n
x
n +
( n + 1) · n!
n!
x n
x
n + 1
| x |
n + 1
n →∞ −→ 0
Since the value of the limit is 0, no matter what real number we choose for x and 0 < 1, the series converges absolutely
for all values of x. ( x ∈ R , −∞ < x < ∞ , (−∞ , ∞)).
Fact : There is always at least one point for which a power series converges: the point x = a at which the series is centred.
