Power Series

The Essentials

A power series is a series that involves some variable, usually \( x \). The general form of a power series centered at 0 is

\[ \Sigma_{n=0}^\infty c_n x^n \]

where \( c_n \) is any function of n. A power series can also be centered at some number \( a \), in which case it will take the form

\[ \Sigma_{n=0}^\infty c_n (x-a)^n \]

Some examples of power series are:

\[ \Sigma_{n=0}^\infty \frac{x^n}{n!} = 1 + \frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6} + ... \]
\[ \Sigma_{n=0}^\infty \frac{(x-2)^n}{(n+1)3^n} = 1 + \frac{x-2}{6} + \frac{(x-2)^2}{27} + \frac{(x-2)^3}{108} + ... \]
\[ \Sigma_{n=0}^\infty 4x^n = 4 + 4x + 4x^2 + 4x^3 + ... \]

The convergence of a power series is dependent on what value x takes. Thus, each power series has an associated interval of convergence, which is the values of x for which the series will converge. Power series also each have a radius of convergence, which is the value \( R \) for which the series will converge if \( |x-a| < R \). The radius of convergence can also be thought of as half of the interval of convergence: if the interval of convergence for a series is \( a < x < b \), the radius of convergence will be \( \frac{b-a}{2} \).

We typically use the ratio test or the root test to determine the convergence of power series. With these tests, we know a series will converge if our computed limit results in \( \rho < 1 \). With a power series, the ratio and root test limits will end up being expressions involving \( x \), which we can then use to solve for the values of \( x \) that make \( \rho < 1 \).

Example

If we want to determine the interval and radius of convergence of the series

\[ \Sigma_{n=1}^\infty \frac{(2x)^n}{n} \]

our first step is to apply the ratio or root test. In this case, both tests will work, but the root test may be simpler since we have n-order exponents in our series:

\[ \rho = \lim\limits_{n\to\infty} \Big |\frac{(2x)^n}{n} \Big |^{\frac{1}{n}} = \lim\limits_{n\to\infty} \Big |\frac{2x}{n^{\frac{1}{n}}} \Big | = |2x| \]

Now we have \( \rho \) as an expression of \( x \). We know that, for the series to converge, \( \rho \) must be less than 1. Thus, the values of \( x \) that will make the series converge must satisfy the equation \( |2x| < 1 \). We can split this equation into two different inequalities and solve them to find the interval of convergence:

\[ |2x| < 1 \longrightarrow 2x < 1, -2x < 1 \longrightarrow x < \frac{1}{2}, x>-\frac{1}{2} \longrightarrow -\frac{1}{2} < x < \frac{1}{2} \]

This tells us that our power series will converge as long as x is between the values \( -\frac{1}{2} \) and \( \frac{1}{2} \). Our radius of convergence is half of this interval: \( R=\frac{\frac{1}{2}-(-\frac{1}{2})}{2}=\frac{1}{2} \).

However, now that we've found our interval of convergence, we still need to figure out if our interval has open or closed bounds. We don't know whether the series will converge at \( x=\frac{1}{2} \) or \( -\frac{1}{2} \) until we plug those values into our series and test for convergence. We'll start with \( x=\frac{1}{2} \):

\[ \Sigma_{n=1}^\infty \frac{(2x)^n}{n} \longrightarrow \Sigma_{n=1}^\infty \frac{(2(\frac{1}{2}))^n}{n} \longrightarrow \Sigma_{n=1}^\infty \frac{(1)^n}{n} \]

Since \( 1^n \) will always equal \( 1 \), this series is equivalent to \( \Sigma_{n=1}^\infty \frac{1}{n} \), which is the harmonic series that is known to diverge.

For \( x=-\frac{1}{2} \):

\[ \Sigma_{n=1}^\infty \frac{(2x)^n}{n} \longrightarrow \Sigma_{n=1}^\infty \frac{(-1)^n}{n} \]

We can apply the alternating series test to this series. Since \( \lim\limits_{n\to\infty} \Sigma_{n=1}^\infty \frac{(-1)^n}{n}=0 \), and \( \frac{(-1)^n}{n} \) is a decreasing sequence, this series converges.

Now that we know whether the series converges or diverges at each boundary value of \( x \), we can write our full interval of convergence as \( \mathbf{-\frac{1}{2} \leq x < \frac{1}{2}} \). Our radius of convergence is still \( \frac{1}{2} \).

Practice

Find the interval and radius of convergence for the following infinite series:

  1. \( \Sigma_{n=1}^\infty \frac{5^n}{n}(3x-1)^{n-1} \)
  2. \( \Sigma_{n=1}^\infty \frac{nx^n}{e^n} \)

Find the radius of convergence for the following infinite series:

  1. \( \Sigma_{n=1}^\infty \frac{(n!)^3x^n}{(3n)!} \)

Solution:

  1. interval of convergence = \( \frac{4}{15}\leq x < \frac{6}{15} \); radius of convergence= \( \frac{1}{15}\)
  2. interval of convergence = \( -e < x < e \); radius of convergence= \( e \)
  3. radius of convergence = \( 27 \)

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