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 $\displaystyle \sum_{k=1}^{\infty} \dfrac{x^k}{k}$.

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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.