How to find the radius and interval of convergence of a Taylor series

The radius of convergence tells us how far we can move from the center of a Taylor series before the series diverges. The interval also tells us which endpoints to include. We’ll find both for the Taylor series of a logarithm centered at 3.

Find the Taylor series about x = 3

Let $f (x) = \ln (2 x)$ for $x > 0$ . The constant term is $f (3) = \ln (6)$ . For $n \geq 1$ , we get the higher derivatives by differentiating $1 / x$ repeatedly.

\begin{array}{l} f^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)!}{x^{n}} \\ \frac{f^{(n)} (3)}{n!} = \frac{{(- 1)}^{n - 1}}{n 3^{n}} \end{array}

The third-degree Taylor polynomial uses the constant term and the first three powers of $x - 3$ .

\begin{array}{l} T_{3} (x) = \ln (6) + \frac{x - 3}{3} \\ - \frac{{(x - 3)}^{2}}{18} + \frac{{(x - 3)}^{3}}{81} \end{array}

For the full series, use the formula for the coefficient and include all positive powers of the distance from the center.

\ln (6) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} {(x - 3)}^{n}}{n 3^{n}}

Use the ratio test to find the radius

Call the term in the sum $a_{n}$ . Away from the center, the ratio of consecutive terms simplifies because one factor of $x - 3$ remains in the numerator and one factor of $3$ remains in the denominator.

\begin{array}{l} | \frac{a_{n + 1}}{a_{n}} | = \frac{| x - 3 |}{3} \cdot \frac{n}{n + 1} \\ L = \lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = \frac{| x - 3 |}{3} \end{array}

When $L < 1$ , the series converges absolutely. When $L > 1$ , it diverges. The center $x = 3$ is included because all the terms in the sum are zero there.

\frac{| x - 3 |}{3} < 1 when 0 < x < 6

The distance from the center to either endpoint is $3$ , so $R = 3$ .

Test each endpoint in the series

At $x = 0$ , the factor ${(x - 3)}^{n}$ becomes ${(- 3)}^{n}$ . Combined with the coefficient, that makes every term in the sum $- 1 / n$ . The negative of the harmonic series diverges, so we exclude $0$ .

\sum_{n = 1}^{\infty} \frac{- 1}{n} diverges

At $x = 6$ , the factor is $3^{n}$ , and the terms alternate. The magnitudes $1 / n$ decrease to zero, so the alternating series test gives convergence. Their absolute values form the harmonic series, so the convergence at $6$ is conditional.

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} converges

The interval of convergence is $(0, 6]$ .

The open circle excludes 0. The filled circle includes 6.

How to find the radius and interval of convergence of a Taylor series

Find the Taylor series about x = 3

Use the ratio test to find the radius

Test each endpoint in the series

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