Interval and Radius of Convergence

This article uses summation notation.

This article uses two-sided limits

The interval of convergence is the open, closed, or semiclosed range of values of \(x\) for which the Taylor series converges to the value of the function; outside the domain, the Taylor series either is undefined or does not relate to the function. The radius of convergence is half the length of the interval; it is also the radius of the circle within the complex plane in which the series converges.

Convergence may be determined by a variety of methods, but the ratio test tends to provide an immediate value \(r\) for the radius of convergence. The interval of convergence may then be determined by testing the value of the series at the endpoints \(-r\) and \(r\).

Find the interval and radius of convergence for the series \( \displaystyle \sum_{k=1}^{\infty} \dfrac{x^k}{k} \).

---

Use the ratio test to note that the series will converge only if \(x\) satisfies

\[ \lim_{k \to \infty} \left| \frac{\hspace{3mm} \frac{x^{k+1}}{k+1}\hspace{3mm} }{\frac{x^k}{k}}\right| < 1 \implies \lim_{k \to \infty} \left| \frac{kx}{k+1}\right| < 1 \implies \left| x \right| < 1. \]

The preliminary interval of convergence is \( -1 < x < 1 \), but the series could possibly converge at the "endpoints" \(x = -1\) and \(x = 1\) as well.

If \( x = 1 \), then the series becomes \( \displaystyle \sum_{k = 1}^{\infty} \dfrac{1}{k} \), which is the harmonic series. This diverges.

If you put in \( x = -1 \), you get \( \displaystyle \sum_{k=1}^{\infty} \dfrac{(-1)^k}{k}\). This is the alternating harmonic series, which converges by the alternating series test.

Therefore, the interval of convergence is \([-1, \, 1)\), and the radius of convergence is \(1\). \(_\square\)

It may help to note that for simple functions, \(1\) and \(\infty\) are common radii of convergence.