Introduction to Power Series: Learn It 2

Convergence of a Power Series

Now we need to address a crucial question: for which values of [latex]x[/latex] does a power series actually converge? Unlike regular series with constant terms, power series behavior depends entirely on the value of [latex]x[/latex]. The series might converge for some [latex]x[/latex]-values and diverge for others.

Every power series has at least one guaranteed convergence point: its center.

For a power series [latex]\displaystyle\sum_{n=0}^{\infty} c_n (x-a)^n[/latex] centered at [latex]x = a[/latex], when we substitute [latex]x = a[/latex]:

[latex]\displaystyle\sum_{n=0}^{\infty} c_n (a-a)^n = c_0 + 0 + 0 + \cdots = c_0[/latex]

The series reduces to just the constant term [latex]c_0[/latex], so it always converges at its center.

Beyond the center, a power series can behave in exactly one of three ways.

convergence theorem for power series

For any power series [latex]\displaystyle\sum_{n=0}^{\infty} c_n (x-a)^n[/latex], exactly one of these occurs:

  1. Converges only at the center: The series converges at [latex]x = a[/latex] and diverges everywhere else
  2. Converges everywhere: The series converges for all real numbers [latex]x[/latex]
  3. Converges in an interval: There exists [latex]R > 0[/latex] such that:
    • The series converges when [latex]|x-a| < R[/latex]
    • The series diverges when [latex]|x-a| > R[/latex]
    • At the boundary points where [latex]|x-a| = R[/latex], the series may converge or diverge

Proof

Suppose that the power series is centered at [latex]a=0[/latex]. (For a series centered at a value of a other than zero, the result follows by letting [latex]y=x-a[/latex] and considering the series [latex]{\displaystyle\sum _{n=1}^{\infty}} {c}_{n} {y}^{n}[/latex].) We must first prove the following fact:

If there exists a real number [latex]d\ne 0[/latex] such that [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{d}^{n}[/latex] converges, then the series [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}[/latex] converges absolutely for all x such that [latex]|x|<|d|[/latex].

Since [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{d}^{n}[/latex] converges, the nth term [latex]{c}_{n}{d}^{n}\to 0[/latex] as [latex]n\to \infty[/latex]. Therefore, there exists an integer N such that [latex]|{c}_{n}{d}^{n}|\le 1[/latex] for all [latex]n\ge N[/latex]. Writing

[latex]|{c}_{n}{x}^{n}|=|{c}_{n}{d}^{n}|{|\frac{x}{d}|}^{n}[/latex],

 

we conclude that, for all [latex]n\ge N[/latex],

[latex]|{c}_{n}{x}^{n}|\le {|\frac{x}{d}|}^{n}[/latex].

 

The series

[latex]\displaystyle\sum _{n=N}^{\infty }{|\frac{x}{d}|}^{n}[/latex]

 

is a geometric series that converges if [latex]|\frac{x}{d}|<1[/latex]. Therefore, by the comparison test, we conclude that [latex]\displaystyle\sum _{n=N}^{\infty }{c}_{n}{x}^{n}[/latex] also converges for [latex]|x|<|d|[/latex]. Since we can add a finite number of terms to a convergent series, we conclude that [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}[/latex] converges for [latex]|x|<|d|[/latex].

With this result, we can now prove the theorem. Consider the series

[latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex]

 

and let S be the set of real numbers for which the series converges. Suppose that the set [latex]S=\left\{0\right\}[/latex]. Then the series falls under case i. Suppose that the set S is the set of all real numbers. Then the series falls under case ii. Suppose that [latex]S\ne \left\{0\right\}[/latex] and S is not the set of real numbers. Then there exists a real number [latex]x*\ne 0[/latex] such that the series does not converge. Thus, the series cannot converge for any x such that [latex]|x|>|x*|[/latex]. Therefore, the set S must be a bounded set, which means that it must have a smallest upper bound. (This fact follows from the Least Upper Bound Property for the real numbers, which is beyond the scope of this text and is covered in real analysis courses.) Call that smallest upper bound R. Since [latex]S\ne \left\{0\right\}[/latex], the number [latex]R < 0[/latex]. Therefore, the series converges for all [latex]x[/latex] such that [latex]|x| < R[/latex], and the series falls into case iii.

[latex]_\blacksquare[/latex]

If a series [latex]\displaystyle\sum_{n=0}^{\infty} c_n (x-a)^n[/latex] falls into the third case and converges in an interval, then we have the most common and interesting situation. The series converges for all [latex]x[/latex] such that [latex]|x-a| < R[/latex] for some [latex]R > 0[/latex], and diverges for all [latex]x[/latex] such that [latex]|x-a| > R[/latex].

What happens at the boundary values where [latex]|x-a| = R[/latex]? The series may converge or diverge — we need to check these endpoints individually.