Convergence of Power Series

A power series is an infinite series

$$a_0 + a_1(x - c) + a_2(x - c)^2 + \cdots = \sum_{n=0}^\infty a_n(x - c)^n.$$

The number c is called the expansion point.

A power series may represent a function $f(x)$ , in the sense that wherever the series converges, it converges to $f(x)$ . There are two issues here:

  1. Where does the series converge?
  2. If the series converges at a point, does it converge to $f(x)$ ?

Example. Consider the series

$$\sum_{n=0}^\infty u^n = 1 + u + u^2 + u^3 + \cdots + u^n + \cdots.$$

Since the terms of the series involve powers of the variable u (i.e. $u - 0$ ), the expansion point is $c = 0$ .

This series represents the function $\dfrac{1}{1 - u}$ for $-1 < u < 1$ . You can see that this is reasonable by dividing 1 by $1 - u$ , or using the the formula for the sum of a geometric series with ratio $r = u$ .

For example, if $u = \dfrac{1}{2}$ ,

$$\dfrac{1}{1 - u} = 2, \quad\hbox{while}\quad 1 + u + u^2 + u^3 + \cdots = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \cdots.$$

Results on geometric series show that the two expressions are equal.

On the other hand, if $u = -1$ ,

$$\dfrac{1}{1 - u} = \dfrac{1}{2}, \quad\hbox{while}\quad 1 + u + u^2 + u^3 + \cdots = 1 - 1 + 1 - 1 + 1 - \cdots.$$

The two expressions aren't equal; in fact, the series on the right diverges, by the Zero Limit Test.

You can use the Ratio Test (and sometimes, the Root Test) to determine the values for which a power series converges. Here are some important facts about the convergence of a power series.

The set of points where the series converges is called the interval of convergence.

Example. The power series

$$a_0 + a_1(x - 5) + a_2(x - 5)^2 + a_3(x - 5)^2 + \cdots$$

is expanded around $c = 5$ . It surely converges at $x = 5$ , since setting $x = 5$ gives $a_0 + 0 + 0 + \cdots = a_0$ .

The series converges on an interval which is symmetric about $a = 5$ . Thus, $-2 < x < 12$ is a possible interval of convergence; $3 < x < 8$ is not.

Suppose you know that $3 < x < 7$ is the largest open interval on which the series converges. Then the series can do anything (in terms of convergence or divergence) at $x = 3$ and $x = 7$ . The interval of convergence could be $3 < x < 7$ (diverges at both ends), $3 \le x \le 7$ (converges at both ends), or $3 \le x < 7$ or $3 < x \le 7$ (converges at one end and diverges at the other).

Example. Determine the interval of convergence for the series $\displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{3^n}$ .

Take absolute values and apply the Ratio Test:

$$\lim_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty} \dfrac{\dfrac{|x|^{n+1}}{3^{n+1}}} {\dfrac{|x|^n}{3^n}} = \lim_{n \to \infty} \dfrac{|x|^{n+1}}{|x|^n}\cdot \dfrac{3^n}{3^{n+1}} = \lim_{n \to \infty} \dfrac{|x|}{3}.$$

The series converges for $\dfrac{|x|}{3} < 1$ . Solving the inequality, I get $|x| < 3$ , or $-3 < x < 3$ . The series diverges for $x < -3$ and for $x > 3$ .

I'll test the endpoints separately.

At $x = 3$ , the series is

$$\sum_{n=0}^{\infty} \dfrac{3^n}{3^n} = \sum_{n=0}^{\infty} 1 = 1 + 1 + 1 + \cdots.$$

The series diverges at $x = 3$ .

At $x = -3$ , the series is

$$\sum_{n=0}^{\infty} \dfrac{(-3)^n}{3^n} = \sum_{n=0}^{\infty} (-1)^n = 1 - 1 + 1 - \cdots.$$

The series diverges at $x = -3$ .

All together, the series diverges for $x \le -3$ and for $x \ge 3$ . It converges for $-3 < x < 3$ .

You would reach the same conclusion using the Root Test:

$$\lim_{n\to \infty} |a_n|^{1/n} = \lim_{n\to \infty} \left(\dfrac{|x|^n}{3^n}\right)^{1/n} = \dfrac{|x|}{3}.$$

The Root Test says that the series converges for $\dfrac{|x|}{3} < 1$ , i.e. for $-3 < x < 3$ , and that it diverges for $x < -3$ and for $x > 3$ . The endpoint check is the same as above.

Example. Determine the interval of convergence for the series $\displaystyle \sum_{n=1}^{\infty} \dfrac{(x - 2)^n}{n \cdot 5^n}$ .

Take absolute values and apply the Ratio Test:

$$\lim_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty} \dfrac{\dfrac{|x - 2|^{n+1}}{(n + 1) \cdot 5^{n+1}}}{\dfrac{|x - 2|^n}{n \cdot 5^n}} = \lim_{n \to \infty} \dfrac{n}{n + 1}\cdot \dfrac{5^n}{5^{n+1}}\cdot \dfrac{|x - 2|^{n+1}}{|x - 2|^n} = \lim_{n \to \infty} \dfrac{1}{5} \dfrac{n}{n + 1} |x - 2| = \dfrac{1}{5} |x - 2|.$$

By the Ratio Test, the series converges (absolutely) for $\dfrac{1}{5} |x - 2| < 1$ , or $-3 < x < 7$ . Likewise, the series diverges for $x < -3$ or for $x > 7$ .

Check the situation at the endpoints. For $x = 7$ , the series becomes $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n}$ . This is the harmonic series, and it diverges.

For $x = -3$ , the series is $\displaystyle \sum_{n=1}^{\infty} (-1)^n \dfrac{1}{n}$ . This is the alternating harmonic series, and it converges by the Alternating Series Test.

To summarize, the series converges absolutely for $-3 < x < 7$ , converges conditionally for $x = -3$ , and diverges for $x < -3$ and for $x \ge 7$ . The interval of convergence is $-3 \le x < 7$ .

Example. Determine the interval of convergence for the series $\displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$ .

Take absolute values and apply the Ratio Test:

$$\lim_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty} \dfrac{\dfrac{|x|^{n+1}}{(n + 1)!}} {\dfrac{|x|^n}{n!}} = \lim_{n \to \infty} \dfrac{|x|^{n+1}}{|x|^n}\cdot \dfrac{n!}{(n + 1)!} = \lim_{n \to \infty} \dfrac{|x|}{n + 1} = 0.$$

The limit is less than 1, independent of the value of x. It follows that the series converges for all x. That is, the interval of convergence is $-\infty < x < +\infty$ .

In fact, this series represents the exponential function:

$$e^x = \sum_{n=0}^{\infty} \dfrac{x^n}{n!}.\quad\halmos$$

Example. Determine the interval of convergence for the series $\displaystyle \sum_{n=0}^{\infty} n^nx^n$ .

Take absolute values and apply the Ratio Test:

$$\lim_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty} \dfrac{(n+1)^{n+1}|x|^{n+1}}{n^n|x|^n} = \lim_{n \to \infty} (n + 1)|x|\cdot \left(\dfrac{n + 1}{n}\right)^n.$$

I'll compute the limit of the last term separately. Let $y = \left(\dfrac{n + 1}{n}\right)^n$ . Then

$$\ln y = \ln \left(\dfrac{n + 1}{n}\right)^n = n \ln \left(\dfrac{n + 1}{n}\right) = n \ln \left(1 + \dfrac{1}{n}\right).$$

By L'Hopital's Rule,

$$\lim_{n \to \infty} \ln y = \lim_{n \to \infty} n \ln \left(1 + \dfrac{1}{n}\right) = \lim_{n \to \infty} \dfrac{\ln \left(1 + \dfrac{1}{n}\right)} {\dfrac{1}{n}} = \lim_{n \to \infty} \dfrac{\left(\dfrac{1}{1 + \dfrac{1}{n}}\right) \left(-\dfrac{1}{n^2}\right)}{\left(-\dfrac{1}{n^2}\right)} = \lim_{n \to \infty} \left(\dfrac{1}{1 + \dfrac{1}{n}}\right) = 1.$$

Hence,

$$\lim_{n \to \infty} y = e^{(\lim_{n\to \infty} \ln y)} = e^1 = e.$$

Therefore,

$$\lim_{n \to \infty} (n + 1)|x|\cdot \left(\dfrac{n + 1}{n}\right)^n = +\infty.$$

This means that the series diverges for all x except $x = 0$ , the point of expansion. At $x = 0$ , the series looks like

$$\sum_{n=0}^{\infty} n^n0^n = \sum_{n=0}^{\infty} 0 = 0,$$

so it certainly converges.

Example. Determine the interval of convergence for the series $\displaystyle \sum_{n=1}^{\infty} \dfrac{2^{2n}x^n}{n^2}$ .

Take absolute values and apply the Ratio Test:

$$\lim_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty} \dfrac{\dfrac{2^{2n+2}|x|^{n+1}}{(n + 1)^2}} {\dfrac{2^{2n}|x|^n}{n^2}} = \lim_{n \to \infty} \dfrac{|x|^{n+1}}{|x|^n}\cdot \dfrac{n^2}{(n + 1)^2}\cdot \dfrac{2^{2n+2}}{2^{2n}} = \lim_{n \to \infty} 4|x|\cdot \dfrac{n^2}{(n + 1)^2} = 4|x|.$$

The series converges for $4|x| < 1$ , i.e. for $-\dfrac{1}{4} < x < \dfrac{1}{4}$ . The series diverges for $x < -\dfrac{1}{4}$ and for $x > \dfrac{1}{4}$ .

I'll test the endpoints separately.

At $x = \dfrac{1}{4}$ , the series is

$$\sum_{n=1}^{\infty} \dfrac{2^{2n}}{4^n\cdot n^2} = \sum_{n=1}^{\infty} \dfrac{1}{n^2}.$$

The series is a p-series with $p = 2 > 1$ , so it converges.

At $x = -\dfrac{1}{4}$ , the series is

$$\sum_{n=1}^{\infty} \dfrac{(-1)^n2^{2n}}{4^n\cdot n^2} = \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^2}.$$

This series converges absolutely, since the absolute value series is $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2}$ , a p-series with $p = 2 > 1$ . Hence, it converges.

All together, the power series converges for $-\dfrac{1}{4} \le x \le \dfrac{1}{4}$ , and diverges for $x < -\dfrac{1}{4}$ and for $x > \dfrac{1}{4}$ .

Example. Determine the interval of convergence for the series $\displaystyle \sum_{n=0}^{\infty} \dfrac{(x + 3)^n}{n + 1}$ .

Take absolute values and apply the Ratio Test:

$$\lim_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty} \dfrac{\dfrac{|x + 3|^{n+1}}{n + 2}} {\dfrac{|x + 3|^n}{n + 1}} = \lim_{n \to \infty} \dfrac{|x + 3|^{n+1}}{|x + 3|^n}\cdot \dfrac{n + 1}{n + 2} = \lim_{n \to \infty} |x + 3|\cdot \dfrac{n + 1}{n + 2} = |x + 3|.$$

The series converges for $|x + 3| < 1$ , i.e. for $-4 < x < -2$ , and diverges for $x < -4$ and for $x > -2$ .

I'll test the endpoints separately.

At $x = -4$ , the series is $\displaystyle \sum_{n=0}^{\infty} \dfrac{(-1)^n}{n + 1}$ . The series alternates. If $f(n) = \dfrac{1}{n + 1}$ , then $f'(n) = -\dfrac{1}{(n + 1)^2} < 0$ , so the terms decrease in absolute value. Finally,

$$\lim_{n \to \infty} \dfrac{1}{n + 1} = 0.$$

Therefore, the series converges, by the Alternating Series Test.

At $x = -2$ , the series is $\displaystyle \sum_{n=0}^{\infty} \dfrac{1}{n + 1}$ . Apply Limit Comparison:

$$\lim_{n\to \infty} \dfrac{\dfrac{1}{n + 1}}{\dfrac{1}{n}} = \lim_{n\to \infty} \dfrac{n}{n + 1} = 1.$$

The limit is a finite positive number. The series $\displaystyle \sum_{n=0}^{\infty} \dfrac{1}{n}$ diverges, because it is harmonic. Therefore, $\displaystyle \sum_{n=0}^{\infty} \dfrac{1}{n + 1}$ , by Limit Comparison.

All together, the power series converges for $-4 \le x < 2$ , and diverges for $x < -4$ and for $x \ge -2$ .

Example. Determine the interval of convergence for the series

$$\dfrac{\ln 3}{3}x^3 + \dfrac{\ln 4}{4}x^4 + \cdots + \dfrac{\ln n}{n}x^n + \cdots.$$

Take absolute values and apply the Ratio Test:

$$\lim_{n \to \infty} \dfrac{|a_{n+1}|}{|a_n|} = \lim_{n \to \infty} \dfrac{\dfrac{\ln (n + 1)}{n + 1}|x|^{n+1}} {\dfrac{\ln n}{n}|x|^n} = \lim_{n \to \infty} \dfrac{\ln (n + 1)}{\ln n}\cdot \dfrac{n + 1}{n}\cdot |x|.$$

By L'Hopital's Rule,

$$\lim_{n \to \infty} \dfrac{\ln (n + 1)}{\ln n} = \lim_{n\to \infty} \dfrac{\dfrac{1}{n + 1}}{\dfrac{1}{n}} = \lim_{n\to \infty} \dfrac{n}{n + 1} = 1.$$

Therefore,

$$\lim_{n \to \infty} \dfrac{\ln (n + 1)}{\ln n}\cdot \dfrac{n + 1}{n}\cdot |x| = |x|.$$

The series converges for $|x| < 1$ , i.e. for $-1 < x < 1$ , and diverges for $x < -1$ and for $x > 1$ .

I'll test the endpoints separately.

At $x = 1$ , the series is $\displaystyle \sum_{n=3}^\infty \dfrac{\ln n}{n}$ . The terms are positive; if $f(n) = \dfrac{\ln n}{n}$ , then

$$f'(n) = -\dfrac{\ln n}{n^2} + \dfrac{1}{n^2} = \dfrac{1 - \ln n}{n^2} < 0 \quad\hbox{for}\quad n \ge 3.$$

Thus, the terms decrease. Apply the Integral Test:

$$\int_3^\infty \dfrac{\ln x}{x}\,dx = \lim_{b\to \infty} \int_3^b \dfrac{\ln x}{x}\,dx = \lim_{b\to \infty} \left[\dfrac{1}{2}(\ln x)^2\right]_3^b = \lim_{b\to \infty} \dfrac{1}{2}\left((\ln b)^2 - (\ln 3)^2\right) = +\infty.$$

Here's the work for the integral:

$$\int \dfrac{\ln x}{x}\,dx = \int \dfrac{u}{x}\cdot x\,du = \int u\,du = \dfrac{1}{2}u^2 + C = \dfrac{1}{2}(\ln x)^2 + C.$$

$$\left[u = \ln x, \quad du = \dfrac{dx}{x}, \quad dx = x\,du\right]$$

Since the integral diverges, the series diverges, by the Integral Test.

At $x = -1$ , the series is $\displaystyle \sum_{n=3}^\infty (-1)^n\dfrac{\ln n}{n}$ . The terms alternate, and the computation above shows that the terms decrease in absolute value. Finally, by L'Hopital's Rule,

$$\lim_{n\to \infty} \dfrac{\ln n}{n} = \lim_{n\to \infty} \dfrac{\dfrac{1}{n}}{1} = 0.$$

By the Alternating Series Test, the series converges.

All together, the series converges for $-1 \le x < 1$ , and diverges for $x < -1$ and for $x \ge 1$ .

Example. Find the function represented by the power series $\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{5^n}(x - 1)^n$ at the points where the series converges.

The series is geometric, so

$$\sum_{n=0}^\infty \dfrac{(-1)^n}{5^n}(x - 1)^n = \dfrac{1}{1 - \left(-\dfrac{1}{5}(x - 1)\right)} = \dfrac{1}{1 + \dfrac{1}{5}(x - 1)} = \dfrac{5}{5 + (x - 1)} = \dfrac{5}{4 + x}.\quad\halmos$$

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