Radius of convergence - another example

Up a level : Power Series
Previous page : Radius of convergence and interval of convergence
Next page : Series - some general rules

Let us have a look at

$f(x) = 1 + \frac{x}{2} + \frac{{{x^2}}}{4} + \frac{{{x^3}}}{8} + \frac{{{x^4}}}{{16}} + ...$

We have

$\left| {{u_k}} \right| = \left| {\frac{{{x^k}}}{{{2^k}}}} \right|$

and thus

$\begin{gathered} \mathop {\lim }\limits_{k \to 0} \left| {\frac{{{u_{k + 1}}}}{{{u_k}}}} \right| = \mathop {\lim }\limits_{k \to 0} \left| {\frac{{\;\frac{{{x^{k + 1}}}}{{{2^{k + 1}}}}\;}}{{\frac{{{x^k}}}{{{2^k}}}}}} \right| = \hfill \\ \quad = \mathop {\lim }\limits_{k \to 0} \left| {\frac{{{x^{k + 1}}}}{{{2^{k + 1}}}} \cdot \frac{{{2^k}}}{{{x^k}}}} \right| = \mathop {\lim }\limits_{k \to 0} \left| {x\frac{1}{2}} \right| = \frac{{\left| x \right|}}{2} < 1 \hfill \\ \end{gathered}$

That means that

$\left| x \right| < 2$

So the radius of convergence is 2.

For the interval of convergence, we need to look at the endpoints. We have

$f(2) = 1 + \frac{2}{2} + \frac{{{2^2}}}{4} + \frac{{{2^3}}}{8} + \frac{{{2^4}}}{{16}} + ...$

and

$f( - 2) = 1 - \frac{2}{2} + \frac{{{2^2}}}{4} - \frac{{{2^3}}}{8} + \frac{{{2^4}}}{{16}} - ...$

None of these are convergent. So the interval is all x such that|x|<2.

The figure below shows the graph of this calculated with 150 terms.

Up a level : Power Series
Previous page : Radius of convergence and interval of convergence
Next page : Series - some general rules

Last modified: Dec 27, 2023 @ 22:52