# Power Series

A **power series** is an infinite series that takes the general form

\[\sum_{n=0}^{\infty} a_n (x-c)^{n} =a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\cdots,\]

where \(a_n\) is a function independent of \(x\) and \(c\) is an arbitrary constant. Thus, each coefficient \((\)e.g. \(a_1, a_2, \ldots)\) can be the result when the number for an \(n^\text{th}\) term is substituted into a given formula. Suppose the \(n^\text{th}\) term for \(a_n\) is equal to \(\frac{n!}{n+1}\). Then the \(5^\text{th}\) term would be \(\frac{5!}{5+1} = \frac{120}{6} = 20\).

Power series is closely linked to the topic of **Taylor series** as it can be used to approximate certain functions about the value \(x = c\).

Any power series can give an approximation about the *center* of the series, denoted by the constant \(c\) above. A power series will converge provided it does not stray too far from this center.

Common problems on power series involve finding the *radius of convergence* and the *Interval of convergence* of a series.

## Common Power Series

One of the most common examples of a power series is that of a **geometric series**. In this case, the value of \(a_n\) remains constant throughout and the value of \(c\) is set to 0. Thus,

\[\sum_{n=0}^{\infty} a x^{n} =a+ax+ax^2+ax^3+\cdots,\]

where each successive term increases by a factor of \(x\).

Another common application of power series is that of **Maclaurin series**, which is a special type of Taylor series with the added stipulation that the center is exclusively about the value \(x=0\). So,

\[f(0) \approx \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = f(0) + f^{(1)}(0)x + \frac{f^{(2)}(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots\]

can be used to approximate any function \(f(x)\) about the value \(x=0\), with the result being more accurate when more terms are part of the summation.

## Radius and Interval of Convergence

In general, a function of the form \( \displaystyle \sum_{n=0}^{\infty} f(n)\) will converge given that \(L < 1\) such that

\[L = \lim_{n \to \infty} \left| \frac{f(n+1)}{f(n)} \right|.\]

This is a standard process and is known as the **ratio test**.

Find the radius of convergence and the interval of convergence of the power series \( \displaystyle \sum_{n=0}^{\infty} \frac{(x-3)^n}{n}.\)

First, perform the ratio test on the function.

\[\begin{align} L &= \lim_{n \to \infty} \left| \frac{f(n+1)}{f(n)} \right| \\ &= \lim_{n \to \infty} \left| \frac{(x-3)^{n+1}}{n+1} \div \frac{(x-3)^n}{n} \right| \\ &= \lim_{n \to \infty} \left| \frac{n(x-3)}{n+1} \right|. \end{align} \]

In order to isolate the function of \(x\), we can simply remove it from the limit, as it is independent of \(n\):

\[\begin{align} L &= \left| x-3\right| \times \lim_{n \to \infty} \left| \frac{n}{n+1} \right| \\ &= \left| x-3\right| \times 1 \\ &= \left| x-3\right|. \end{align}\]

Given that \(L < 1\) in order for the series to converge, it must be that \(\left| x-3 \right| < 1\) when it converges. When you have an expression of the form \(\left| x + a \right| < R\) (so that \(a\) is a member of the reals), the

radius of convergenceis the value of \(R\). So, the radius of convergence in the example is 1.The

Interval of convergence, on the other hand, is the set of all values of \(x\) for which the series converges. Using the above inequality, it must be that\[\begin{align} \left| x-3 \right| &< 1\\ -1 < x-3 &< 1\\ 2 < x &< 4. \end{align}\]

However, it's not as simple as it looks at first glance. We also need to check the boundary values of the interval to check whether the series converges for these values. So, for \(x=2,\) we have

\[ \displaystyle \sum_{n=0}^{\infty} \frac{(2-3)^n}{n} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n},\]

which, using the

alternating series test, converges. Now, for \(x=4,\)\[ \displaystyle \sum_{n=0}^{\infty} \frac{(4-3)^n}{n} = \sum_{n=0}^{\infty} \frac{1^n}{n} = \sum_{n=0}^{\infty} \frac{1}{n},\]

which, as a standard result, does not converge. Thus the Interval of convergence for \( \displaystyle \sum_{n=0}^{\infty} \frac{(x-3)^n}{n}\) is

\(2 \leq x < 4\). \(_\square\)

Find the radius of convergence of \( \displaystyle \sum_{n=0}^{\infty} n! \cdot x^n.\)

Perform the ratio test on the function:

\[\begin{align} L &= \lim_{n \to \infty} \left| \frac{(n+1)! \cdot x^{n+1}}{n! \cdot x^n} \right| \\ &= \lim_{n \to \infty} \left| x(n+1) \right|\\ &= \left| x \right| \cdot \lim_{n \to \infty} \left| (n+1) \right|. \end{align}\]

This example varies from the above example as the limit does not converge to 1 as \(n \to \infty\). For any non-zero value of \(x\), \(L\) will always be equal to \(\infty\). The only value that \(x\) can take to differ from this is \(0\). When this is true, \(L = 0,\) which also satisfies the condition for convergence \((L < 1).\) Thus, the radius of convergence for the series is 0 and the Interval of convergence is \(x=0\). \(_\square\)

Find the radius of convergence of \( \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{2^{2n}(n!)^2}. \)

Perform the ratio test on the function:

\[\begin{align} L &= \lim_{n \to \infty} \left| \frac{(-1)^{n+1} x^{2n+2} (n+1)}{2^{2n+2}((n+1)!)^2} \div \frac{(-1)^n x^{2n}}{2^{2n}(n!)^2} \right| \\ &= \lim_{n \to \infty} \left| - \frac{x^2}{4n(n+1)} \right| \\ &= \lim_{n \to \infty} \left| \frac{x^2}{4n(n+1)} \right| \\ &= \left|x^2\right| \cdot \lim_{n \to \infty} \left| \frac{1}{4n(n+1)} \right|. \end{align} \]

The limit in this scenario approaches 0 and thus this result will hold for all \(x\). This means that \(x\) can take any value and \(L = 0 < 1,\) which means that \( \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{2^{2n}(n!)^2} \) will converge for all \(x\). The radius of convergence in this instance is, therefore, \(\infty\) and the Interval of convergence is \(-\infty < x < \infty\). \(_\square\)

Examples 1 and 2 taken from Calculus 7E Early Transcendentals by James Stewart.

Special Thanks to Thomas Shelly, math professor at Mount Holyoke College.