# Radius of convergence and interval of convergence

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Previous page : Logarithms
Next page : Radius of convergence - another example We can use the ratio test for series to find the values for which the series converges.

We have that a series $\sum\limits_{k = 0}^\infty {{u_k}}$

converges if $\mathop {\lim }\limits_{k \to 0} \left| {\frac{{{u_{k + 1}}}}{{{u_k}}}} \right| < 1$

It basically means that a series converges if its terms grows slower than the terms of a convergent geometric series (at least after a number of terms). If the limit is larger than one the terms gets bigger and bigger, and the series will diverge, and if the limit is 1, then the test is inconclusive.

If we for example look at $f(x) = \frac{1}{{1 + x}} = 1 - x + {x^2} - {x^3} + ...$

So $\left| {{u_k}} \right| = {x^k}$

and we thus have that $\mathop {\lim }\limits_{k \to 0} \left| {\frac{{{u_{k + 1}}}}{{{u_k}}}} \right| = \mathop {\lim }\limits_{k \to 0} \left| {\frac{{{x^{k + 1}}}}{{{x^k}}}} \right| = \mathop {\lim }\limits_{k \to 0} \left| x \right|$

This should be less than 1 for the series to converge, and we thus have that the series converges when $\left| x \right| < 1$

The radius of convergence is thus 1.

We must check the endpoints of the interval separately. In this case it is easy to see that neither f(1) nor f(-1) will converge, and the interval of convergence will this be (–1,1) (or ]–1,1[ or {x| –1<x<1}, depending on the way of writing it).

Logarithms

For $\ln (1 + x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + ...$

we have $\left| {{u_k}} \right| = \left| {\frac{{{x^k}}}{k}} \right|$

and thus $\begin{gathered} \mathop {\lim }\limits_{k \to 0} \left| {\frac{{{u_{k + 1}}}}{{{u_k}}}} \right| = \mathop {\lim }\limits_{k \to 0} \left| {\frac{{\;\frac{{{x^{k + 1}}}}{{k + 1}}\;}}{{\frac{{{x^k}}}{k}}}} \right| = \hfill \\ \quad = \mathop {\lim }\limits_{k \to 0} \left| {\frac{{{x^{k + 1}}}}{{k + 1}} \cdot \frac{k}{{{x^k}}}} \right| = \mathop {\lim }\limits_{k \to 0} \left| {x\frac{k}{{k + 1}}} \right| \hfill \\ \quad = \mathop {\lim }\limits_{k \to 0} \left| x \right|\mathop {\lim }\limits_{k \to 0} \left| {\frac{1}{{1 + 1/k}}} \right| = \left| x \right| \hfill \\ \end{gathered}$

We yet again get that the series converges when $\left| x \right| < 1$

So the radius of convergence is thus 1.

For the endpoints we have $\ln (1 + 1) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ...$

that indeed converges to ln(2). But for the other endpoint we have ln(1-1)=ln(0) that is undefined, and we also have that $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...$

converges. So the interval of convergence is (1,–1] or ]1,1].

The exponential function

We have ${e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... = \sum\limits_{k = 0}^\infty {\frac{{{x^k}}}{{k!}}}$

So $\left| {{u_k}} \right| = \frac{{{x^k}}}{{k!}}$

and $\begin{gathered} \mathop {\lim }\limits_{k \to 0} \left| {\frac{{{u_{k + 1}}}}{{{u_k}}}} \right| = \mathop {\lim }\limits_{k \to 0} \left| {\frac{{\;\frac{{{x^{k + 1}}}}{{(k + 1)!}}\;}}{{\frac{{{x^k}}}{{k!}}}}} \right| = \hfill \\ \quad = \mathop {\lim }\limits_{k \to 0} \left| {\frac{{{x^{k + 1}}}}{{(k + 1)!}} \cdot \frac{{k!}}{{{x^k}}}} \right| = \mathop {\lim }\limits_{k \to 0} \left| {x\frac{1}{{k + 1}}} \right| = 0 \hfill \\ \end{gathered}$

This is less than one, so the series will converge for any value of x. It will thus have an infinite radius of convergence. As you can verify this is true for the sine and cos functions too. Up a level : Power Series
Previous page : Logarithms
Next page : Radius of convergence - another example Last modified: Mar 3, 2019 @ 20:04