- 9.1 | Sequences
- 9.2 | Series and Convergence
- 9.3 | The Integral Test and p-Series
- 9.4 | Comparisons of Series
- 9.5 | Alternating Series
- 9.6 | The Ratio and Root Tests
- 9.7 | Taylor Polynomials and Approximations
- 9.8 | Power Series
- 9.10 | Taylor and Maclaurin Series

$$

\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

$$

is called a

**power series in \((x-c)\)**or a

**power series centered at \(c\)**or a

**power series about \(c\)**

$$

\sum_{n=0}^{\infty} a_n (x-c)^n

$$

has interval of convergence \((A,B), [A,B),\) or \((A,B]\) then the radius of convergence is \(\frac{B-A}{2}\).

$$

f(x) = \sum_{n=0}^{\infty} a_n (x-c)^n

$$

is differentiable (and therefore continuous) on the interval \((a -R, a+R)\) and

- \( f'(x) = \sum_{n=1}^{\infty} n a_n (x-c)^{n-1} \)
- \(\displaystyle \int f(x)\; dx = C + \sum_{n=0}^{\infty} a_n \frac{(x-c)^{n+1}}{n+1} \)

The radii of convergence of the power series above are both \(R\).

$$

\sum_{n = 1}^{\infty} x^{n-1}

$$

represents the function

$$

f(x) = \frac{1}{1-x}

$$

for \(|x|<1\). Replacing a power series with its common function representation can serve as a means to evaluate the sum of a power series for a particular value of \(x\) within its interval of convergence. Conversely, replacing a common function with a power series can be useful when integrating a function whose antiderivative is unknown or difficult to obtain. The examples that follow demonstrate how to calculate the interval of convergence and/or radius of convergence of a given power series. The interval of convergence plays an important role in establishing the values of \(x\) for which a power series is equal to its common function representation. Moreover, the interval of convergence is necessary to establish the values of \(x\) for which the derivative or integral of a power series is equal to the power series obtained by differentiating or integrating term-by-term (more on this in section 8.6). Again, in this section we concern ourselves with simply finding the interval of convergence and/or radius of convergence of a given power series.

$$

\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}}

$$

Apply the ratio test:

$$

\require{cancel}

\begin{aligned}

\lim_{n \rightarrow \infty} \left| \frac{\frac{x^{n+1}}{\sqrt{n+1}}}{\frac{x^n}{\sqrt{n}}} \right| & =

\lim_{n \rightarrow \infty} \left| \frac{x^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{x^n}\right| \\

& =

\lim_{n \rightarrow \infty} \left| \frac{x^{n+1}}{\sqrt{n+1}} \frac{\sqrt{n}}{ x^n} \right| \\

& =

\lim_{n \rightarrow \infty} \left| \frac{\bcancel{x^{n}} x}{\sqrt{n+1}} \frac{\sqrt{n}}{ \bcancel{x^n}} \right| \\

& =

\lim_{n \rightarrow \infty} \left| \frac{x \sqrt{n}}{\sqrt{n+1}}\right| \\

& =

\lim_{n \rightarrow \infty} \left(|x| \cdot \left| \frac{ \sqrt{n}}{\sqrt{n+1}}\right|\right) \\

& =

|x| \cdot \lim_{n \rightarrow \infty} \frac{ \sqrt{n}}{\sqrt{n+1}} = |x|

\end{aligned}

$$

So

$$

\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n} \right| = |x|

$$

It suffices to impose that \(|x|<1\) in order for the series to converge. That is, for any \(x\) in \((-1,1)\) the series converges.
To find out whether or not to include both endpoints, one of the endpoints, or none of the endpoints we test the series for convergence at the endpoints:
If \(x = 1\) then
$$
\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}
$$
which diverges by the \(p\)-series test.
If \(x = -1\) then
$$
\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}
$$
which converges by the Alternating Series Test.
Thus the interval of convergence is \([-1,1)\) and the radius of convergence is \(\frac{1-(-1)}{2} = 1\).

$$

\sum_{n=1}^{\infty} \frac{(3x-2)^n}{n3^n}

$$

Apply the ratio test:

$$

\require{cancel}

\begin{aligned}

\lim_{n \rightarrow \infty} \left| \frac{\frac{(3x-2)^{n+1}}{(n+1)3^{n+1}}}{\frac{(3x-2)^n}{n3^n}} \right| & =

\lim_{n \rightarrow \infty} \left| \frac{(3x-2)^{n+1}}{(n+1)3^{n+1}}\frac{n3^n}{(3x-2)^n}\right| \\

& =

\lim_{n \rightarrow \infty} \left| \frac{(3x-2)^n(3x-2)}{(n+1)3^n 3}\frac{n3^n}{(3x-2)^n}\right| \\

& =

\lim_{n \rightarrow \infty} \left| \frac{\bcancel{(3x-2)^n}(3x-2)}{(n+1)\bcancel{3^n} 3}\frac{n \bcancel{3^n}}{\bcancel{(3x-2)^n}}\right| \\

& =

\lim_{n \rightarrow \infty} \left| \frac{(3x-2)n}{3(n+1)}\right| \\

& =

\lim_{n \rightarrow \infty} \left( \left|\frac{3x-2}{3} \right| \cdot \left| \frac{n}{n+1}\right| \right) \\

& =

\left|\frac{3x-2}{3} \right| \lim_{n \rightarrow \infty} \cdot \left| \frac{n}{n+1}\right| = \left|\frac{3x-2}{3} \right| \\

\end{aligned}

$$

So

$$

\lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n} \right| = \left|\frac{3x-2}{3} \right|

$$

It suffices to impose that \(\left|\frac{3x-2}{3} \right|<1\) in order for the series to converge. That is, for any \(x\) in \((-1/3,5/3)\) the series converges.
To find out whether or not to include both endpoints, one of the endpoints, or none of the endpoints we test the series for convergence at the endpoints:
If \(x = 5/3\) then
$$
\sum_{n=1}^{\infty} \frac{(3x-2)^n}{n3^n} = \sum_{n=1}^{\infty} \frac{1}{n3^n}
$$
which converges by the Direct Comparison Test with comparison series \(\sum \frac{1}{n^2} \).
If \(x = -1/3\) then
$$
\sum_{n=1}^{\infty} \frac{(3x-2)^n}{n3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n3^n}
$$
which converges by the Alternating Series Test.
Thus the interval of convergence is \([-1/3,5/3]\) and the radius of convergence is \(\frac{5/3-(-1/3)}{2} = 1\).