The radius of convergence of a power series is infinite when the series converges for all real or complex numbers. This means the power series can be used to define a function that is analytic everywhere in its domain.
What is a Power Series?
A power series is an infinite series of the form∑n=0∞cn(x−a)n=c0+c1(x−a)+c2(x−a)2+…sum from n equals 0 to infinity of c sub n open paren x minus a close paren to the n-th power equals c sub 0 plus c sub 1 open paren x minus a close paren plus c sub 2 open paren x minus a close paren squared plus …
∞𝑛=0𝑐𝑛(𝑥−𝑎)𝑛=𝑐0+𝑐1(𝑥−𝑎)+𝑐2(𝑥−𝑎)2+…
where xx
𝑥
is a variable, aa
𝑎
is a constant called the center of the series, and cnc sub n
𝑐𝑛
are the coefficients. The series may converge for some values of xx
𝑥
and diverge for others.
Radius of Convergence
The radius of convergence, denoted by Rcap R
𝑅
, is a non-negative number (or infinity) that describes the interval or disk in which the series converges.
-
If Rcap R
𝑅
is a finite positive number, the series converges for all xx
𝑥
such that |x−a|<Rthe absolute value of x minus a end-absolute-value is less than cap R
|𝑥−𝑎|<𝑅
and diverges for all xx
𝑥
such that |x−a|>Rthe absolute value of x minus a end-absolute-value is greater than cap R
|𝑥−𝑎|>𝑅
. The interval of convergence is (a−R,a+R)open paren a minus cap R comma a plus cap R close paren
(𝑎−𝑅,𝑎+𝑅)
, and the behavior at the endpoints x=a±Rx equals a plus or minus cap R
𝑥=𝑎±𝑅
must be checked separately.
-
If R=0cap R equals 0
𝑅=0
, the series converges only at its center, x=ax equals a
𝑥=𝑎
.
-
If R=∞cap R equals infinity
𝑅=∞
, the series converges for all real numbers xx
𝑥
, which can be expressed as |x−a|<∞the absolute value of x minus a end-absolute-value is less than infinity
|𝑥−𝑎|<∞
. The interval of convergence is (−∞,∞)open paren negative infinity comma infinity close paren
(−∞,∞)
.
Calculating the Radius of Convergence
The radius of convergence is most commonly found using the Ratio Test or the Root Test.
Ratio Test
The Ratio Test states that a series ∑ansum of a sub n
𝑎𝑛
converges absolutely if the limit of the absolute ratio of consecutive terms is less than 1. For a power series, this means finding the limitL=limn→∞|cn+1(x−a)n+1cn(x−a)n|=limn→∞|cn+1cn||x−a|cap L equals limit over n right arrow infinity of the absolute value of the fraction with numerator c sub n plus 1 end-sub open paren x minus a close paren raised to the n plus 1 power and denominator c sub n open paren x minus a close paren to the n-th power end-fraction end-absolute-value equals limit over n right arrow infinity of the absolute value of the fraction with numerator c sub n plus 1 end-sub and denominator c sub n end-fraction end-absolute-value the absolute value of x minus a end-absolute-value
𝐿=lim𝑛→∞𝑐𝑛+1(𝑥−𝑎)𝑛+1𝑐𝑛(𝑥−𝑎)𝑛=lim𝑛→∞𝑐𝑛+1𝑐𝑛|𝑥−𝑎|
The series converges if L<1cap L is less than 1
𝐿<1
, which gives |x−a|<1limn→∞|cn+1cn|the absolute value of x minus a end-absolute-value is less than the fraction with numerator 1 and denominator limit over n right arrow infinity of the absolute value of the fraction with numerator c sub n plus 1 end-sub and denominator c sub n end-fraction end-absolute-value end-fraction
|𝑥−𝑎|<1lim𝑛→∞𝑐𝑛+1𝑐𝑛
. The radius of convergence is thereforeR=limn→∞|cncn+1|cap R equals limit over n right arrow infinity of the absolute value of the fraction with numerator c sub n and denominator c sub n plus 1 end-sub end-fraction end-absolute-value
𝑅=lim𝑛→∞𝑐𝑛𝑐𝑛+1
If the limit of the ratio of coefficients is zero, i.e., limn→∞|cn+1cn|=0limit over n right arrow infinity of the absolute value of the fraction with numerator c sub n plus 1 end-sub and denominator c sub n end-fraction end-absolute-value equals 0
lim𝑛→∞𝑐𝑛+1𝑐𝑛=0
, then R=∞cap R equals infinity
𝑅=∞
.
Root Test
The Root Test states that a series ∑ansum of a sub n
𝑎𝑛
converges absolutely if the limit of the nn
𝑛
-th root of the absolute value of the terms is less than 1. For a power series, this means finding the limitL=limn→∞|cn(x−a)n|n=limn→∞|cn|n|x−a|cap L equals limit over n right arrow infinity of the n-th root of the absolute value of c sub n open paren x minus a close paren to the n-th power end-absolute-value end-root equals limit over n right arrow infinity of the n-th root of the absolute value of c sub n end-absolute-value end-root the absolute value of x minus a end-absolute-value
𝐿=lim𝑛→∞|𝑐𝑛(𝑥−𝑎)𝑛|𝑛√=lim𝑛→∞|𝑐𝑛|𝑛√|𝑥−𝑎|
The series converges if L<1cap L is less than 1
𝐿<1
, which gives |x−a|<1limn→∞|cn|nthe absolute value of x minus a end-absolute-value is less than the fraction with numerator 1 and denominator limit over n right arrow infinity of the n-th root of the absolute value of c sub n end-absolute-value end-root end-fraction
|𝑥−𝑎|<1lim𝑛→∞|𝑐𝑛|𝑛√
. The radius of convergence isR=1limn→∞|cn|ncap R equals the fraction with numerator 1 and denominator limit over n right arrow infinity of the n-th root of the absolute value of c sub n end-absolute-value end-root end-fraction
𝑅=1lim𝑛→∞|𝑐𝑛|𝑛√
If limn→∞|cn|n=0limit over n right arrow infinity of the n-th root of the absolute value of c sub n end-absolute-value end-root equals 0
lim𝑛→∞|𝑐𝑛|𝑛√=0
, then R=∞cap R equals infinity
𝑅=∞
.
Example: Infinite Radius of Convergence
Consider the power series for the exponential function, centered at a=0a equals 0
𝑎=0
:ex=∑n=0∞xnn!=1+x+x22!+x33!+…e to the x-th power equals sum from n equals 0 to infinity of the fraction with numerator x to the n-th power and denominator n exclamation mark end-fraction equals 1 plus x plus the fraction with numerator x squared and denominator 2 exclamation mark end-fraction plus the fraction with numerator x cubed and denominator 3 exclamation mark end-fraction plus …
𝑒𝑥=∞𝑛=0𝑥𝑛𝑛!=1+𝑥+𝑥22!+𝑥33!+…
Here, the coefficients are cn=1n!c sub n equals the fraction with numerator 1 and denominator n exclamation mark end-fraction
𝑐𝑛=1𝑛!
. Let's use the Ratio Test to find the radius of convergence.|cn+1cn|=|1/(n+1)!1/n!|=|n!(n+1)!|=1n+1the absolute value of the fraction with numerator c sub n plus 1 end-sub and denominator c sub n end-fraction end-absolute-value equals the absolute value of the fraction with numerator 1 / open paren n plus 1 close paren exclamation mark and denominator 1 / n exclamation mark end-fraction end-absolute-value equals the absolute value of the fraction with numerator n exclamation mark and denominator open paren n plus 1 close paren exclamation mark end-fraction end-absolute-value equals the fraction with numerator 1 and denominator n plus 1 end-fraction
𝑐𝑛+1𝑐𝑛=1/(𝑛+1)!1/𝑛!=𝑛!(𝑛+1)!=1𝑛+1
Now, we take the limit as n→∞n right arrow infinity
𝑛→∞
:limn→∞|cn+1cn|=limn→∞1n+1=0limit over n right arrow infinity of the absolute value of the fraction with numerator c sub n plus 1 end-sub and denominator c sub n end-fraction end-absolute-value equals limit over n right arrow infinity of the fraction with numerator 1 and denominator n plus 1 end-fraction equals 0
lim𝑛→∞𝑐𝑛+1𝑐𝑛=lim𝑛→∞1𝑛+1=0
Since this limit is 0, the radius of convergence is R=10=∞cap R equals 1 over 0 end-fraction equals infinity
𝑅=10=∞
. This confirms that the power series for exe to the x-th power
𝑒𝑥
converges for all real numbers xx
𝑥
.
Why is an Infinite Radius of Convergence Important?
A power series with an infinite radius of convergence is a powerful tool because it represents a function that is "well-behaved" everywhere. Such functions are called entire functions in complex analysis. Examples include exe to the x-th power
𝑒𝑥
, sin(x)sine x
sin(𝑥)
, and cos(x)cosine x
cos(𝑥)
. For these functions, their power series representation is valid globally, making them predictable and useful for a wide range of applications, from solving differential equations to performing complex calculations.
Show all