Power Series and Radius of Convergence

When we deal with power series, the radius of convergence becomes a very important concept. But what exactly is the radius of convergence? Let''s find out:

The radius of convergence defined

We know that a series is "convergent" if it approaches a particular finite value. In other words, a convergent series has a sequence of partial sums that approaches a limit as n tends to infinity. We have probably examples of this before -- such as the way certain series approach an asymptote while never actually intersecting with it.

If the series does not show this tendency, then we call it "divergent."

Working with convergent series

How do we know whether this series is convergent:

_∞∑(−1)ⁿ^{n = 0}

We can start by considering the first few partial sums:

s_0 = 1s_1 = 1-1 = 0
s_2 = 1-1+1 =1
s_4 = 1-1+1-1 = 0

We''re not seeing any real tendency toward a limit. The sums are not approaching any number. So we can say that this power series is divergent.

Recall that a power series takes the following form:

_∞Σc_n(x-a)ⁿ^{n = 0}

In this expression, a and c_n are numbers. We call the c_n values "coefficients of the series."

The most important thing to remember is that the values of x determine whether a series is divergent or convergent.

We know that whenever x=a, the power series converges.

So what exactly is the radius of convergence? We can envision this as the radius of a "disc" in which the series converges.

If we know that an interval of convergence is the largest in the entire series, we call this (and only this) the interval of convergence. It is centered at the center of the power series. If we need to find the radius of convergence, we simply take half of the interval. Note that this value could be non-negative, real, or even infinite.

If we know that the radius of convergence is centered at a, then it also converges for the real values of x, such that:

|x-a|

It also diverges for:

|x-a| > r

The series converges in the interval

a-r

a+r

And:
The series diverges for

a-r

and

x > a+r

But how do we find the radius of convergence?

One method is the ratio test:

_∞∑c_n(x−a)ⁿⁿ⁼⁰

This series converges for x such that:

lim|C_n+1(x−a)ⁿ⁺¹÷C_n(x−a)ⁿ|

1^n→∞

Another method is the root test:

_∞∑c_n(x−a)ⁿⁿ⁼⁰

This series converges for x such that:

lim|(C_n(x−a)ⁿ)^1/n|

1^n→∞

Practicing our skills

Now let''s use our knowledge to find the radius of convergence for a given power series. Consider the following power series:

_∞∑(x−1)ⁿ÷2ⁿⁿ⁼⁰

We know that this is a power series centered at x = 1. Let''s use the ratio test to find our interval of convergence:

L =|((x-1)ⁿ⁺¹/2ⁿ⁺¹÷(x-1)ⁿ/2ⁿ)|

This leaves us with:

|((x-1)ⁿ⁺¹/2ⁿ⁺¹)*(2ⁿ/(x-1)ⁿ|

We can simplify this as:

|(x-1)/2|

Now we can take the limits:

lim|(x-1)/2| = |(x-1)/2|^n→∞

The series converges for x such that

|(x-1)/2|

In other words:

-2

x-1

|x-1|

Now we know that the radius of convergence is 2, and the interval of convergence is (-1, 3).

Topics related to the Power Series and Radius of Convergence

Infinity

Sigma Notation of a Series

Sum of the First n Terms of an Arithmetic Sequence

Flashcards covering the Power Series and Radius of Convergence

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AP Calculus BC Flashcards

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