# Power Series

So far, everything we've done with series has dealt with series of constants, where we add up an infinite list of constants and see if the result converges or diverges. A power series, on the other hand, has a variable in it; you can think of it as an infinite polynomial. This series may converge for some values of $x$ and diverge for others. In fact, that's the big question with power series: what values of $x$ make it convergent?

The reason we're interested in power series is that they are necessary for what we'll do in the next section, when we talk about Taylor Series, which are a way to rewrite transcendental functions as power functions (again, essentially polynomials), which are simpler to handle.

# Introduction

A power series is of the form $\sum_{n=0}^{\infty} c_n x^n = c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+\ldots$ where $x$ is a variable.

For example, the following is a power series: $1+\dfrac{1}{2}x+\dfrac{1}{3}x^2+\dfrac{1}{4}x^3+\dfrac{1}{5}x^4+\ldots = \sum_{n=0}^{\infty} \dfrac{1}{n+1}x^n.$

Again, for each value of $x,$ this becomes a series of constants that may converge or diverge, and we'd like to know exactly what $x$ values make it converge. The range of $x$ values for which a power series converges is called the interval of convergence.

### General Form of a Power Series

More generally, $\sum_{n=0}^{\infty}c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+c_3(x−a)^3+c_4(x−a)^4+\ldots$ is called a power series centered at $a$ (in the formula at the beginning of the section, $a=0$). The reason it is said to be centered at $a$ is that when we find the interval of convergence, we'll see that the center of this interval is always $a.$

### Note:

Every power series converges at its center: if $x=a,$ the power series looks like $\sum_{n=0}^{\infty}c_n(a−a)^n=c_0+c_1(a−a)+c_2(a−a)^2+c_3(a−a)^3+\ldots=c_0.$

### Interval of Convergence: Use the Ratio Test

To find the interval of convergence, we'll always use the Ratio Test.

## Interval of Convergence

Find the interval of convergence for $\sum_{n=0}^{\infty} \dfrac{1}{n}(x−3)^n.$

### Solution

First of all, notice that this power series is centered at $3,$ so we know that the interval of convergence will be centered there as well; the interval will be symmetric around that point. Again, to find this interval, use the Ratio Test: \begin{aligned} p &= \lim_{n \to \infty} \left|\dfrac{\dfrac{(x-3)^{n+1}}{n+1}}{\dfrac{(x-3)^n}{n}}\right|\ &= \lim_{n \to \infty} \left|\dfrac{(x-3)^{n+1}}{n+1} \cdot \dfrac{n}{(x-3)^n}\right|\ &= \lim_{n \to \infty} \left|(x-3) \cdot \dfrac{n}{n+1}\right| \end{aligned}

Now, in order for this series to be convergent, the Ratio Test says that this limit must be strictly less than $1$: $\lim_{n \to \infty} \left|(x−3) \cdot \dfrac{n}{n+1}\right|<1$

Notice that we can pull $x−3$ out of the limit (being careful to keep the absolute value signs on it), since it doesn't contain any $n$'s. Then we can evaluate the limit that remains, and we get the following:

\begin{aligned} \lim_{n \to \infty} \left|(x−3) \cdot \dfrac{n}{n+1}\right| &< 1\ |x-3| \lim_{n \to \infty} \left|\dfrac{n}{n+1}\right| &< 1\ |x-3| &< 1 \end{aligned}

It may have been a while since you've solved inequalities like this, but here's the deal: $|a−b|$ means the distance between $a$ and $b$ (if I told you I was at mile marker $18$ and you were at mile marker $25,$ you'd subtract them to find the distance between us, and if you subtracted in the wrong order, you'd just discard the negative sign), so $|x−3|$ is the distance between $x$ and $3.$ This distance must be less than $1,$ so $x$ can't go any lower than $2$ or any higher than $4$: $−2 That's the interval of convergence, with one caveat: we know that this series converges whenever $x$ is in that range, and we know that the series diverges when $x$ is less than $2$ or greater than $4.$ But what about when $x=2$ or $x=4$? In that case, the Ratio Test is inconclusive, remember, so we have to test the endpoints separately.

• When $x=2$: $\sum_{n=0}^{\infty} (−1)^n \dfrac{1}{n}$ is convergent by the Alternating Series Test
• When $x=4$: $\sum_{n=0}^{\infty} \dfrac{1}{n}$ is divergent; this is the harmonic series

This gives us the final answer for the interval of convergence: $\ans{2 \leq x < 4 = [2,4)}$

## Three Possibilities for the IoC

Remember that every power series converges at its center. There are therefore three possibilities:

• The series converges only when $x=a.$
• The series converges for all $x$ values; in this case, the interval of convergence is $(−\infty,\infty).$
• The series converges for some finite interval that is centered at $x=a.$

In this case, the endpoints must be checked separately.

## Interval of Convergence

Find the interval of convergence for $\sum_{n=0}^{\infty} \dfrac{(−1)^n\ n}{4^n} (x+3)^n.$

### Solution

This one is centered at $x=−3,$ so we know that the interval of convergence will be symmetric about that point. Set up the Ratio Test: \begin{aligned} p &= \lim_{n \to \infty} \left|\dfrac{\dfrac{(−1)^{n+1} (n+1) (x+3)^{n+1}}{4^{n+1}}}{\dfrac{(−1)^n (n) (x+3)^n}{4^n}}\right|\ &= \lim_{n \to \infty} \left|\dfrac{(−1)^{n+1} (n+1) (x+3)^{n+1}}{4^{n+1}} \cdot \dfrac{4^n}{(−1)^n (n) (x+3)^n}\right|\ &= \lim_{n \to \infty} \left|(x+3)\left(-\dfrac{1}{4}\right) \cdot \dfrac{n+1}{n}\right| \end{aligned}

Now, in order for this series to be convergent, the Ratio Test says that this limit must be strictly less than $1$: \begin{aligned} \lim_{n \to \infty} \left|(x+3)\left(-\dfrac{1}{4}\right) \cdot \dfrac{n+1}{n}\right| &< 1\ |x+3| \lim_{n \to \infty} \left|\left(-\dfrac{1}{4}\right) \cdot \dfrac{n+1}{n}\right| &< 1\ \dfrac{1}{4} |x+3| &< 1\ |x+3| &< 4 \end{aligned}

Now, $|x+3|=|x−(−3)|$ is the distance between $x$ and $−3,$ and this distance must be less than $4$: $−7

Test the endpoints:

• When $x=−7$: $\sum_{n=0}^{\infty} \dfrac{(−1)^n\ n}{4^n}\ (−4)^n = \sum_{n=0}^{\infty} n$ is divergent (pretty clearly by the Divergence Test; it is also a p-series with $p=−1$
• When $x=1$: $\sum_{n=0}^{\infty} \dfrac{(−1)^n\ n}{4^n}\ (4)^n = \sum_{n=0}^{\infty} (−1)^n\ n$ is divergent by the Divergence Test

This gives us the final answer for the interval of convergence: $\ans{−7