Convergency test for power series

 

POWER SERIES

In mathematics, a power series (in one variable) is an infinite series of the form

where a_n represents the coefficient of the nth term and c is a constant. Power series are useful in mathematical analysis, where they arise as Taylor series of infinitely differentiable functions. In fact, Borel's theorem implies that every power series is the Taylor series of some smooth function.

In many situations, c (the center of the series) is equal to zero, for instance when considering a Maclaurin series. In such cases, the power series takes the simpler form

Radius of convergence

 is convergent for some values of the variable x

if

 |x – c| < r     ,  converges

 |x – c| > r     ,  diverges  

 |x – c| = r      , further investigation required

 r is called the radius of convergence of the power series; in general it is given as

or 

ABSOLUTE CONVERGENCE

A sum of real numbers or complex numbers $\sum _{�=0}^{\mathrm{\infty }}{�}_{�}$ is absolutely convergent if the sum of the absolute values of the terms $\sum _{�=0}^{\mathrm{\infty }}|{�}_{�}|$ converges.

PROOF

First notice that $\left|a_n\right|$ is either $a_n$ or it is $-a_n$ depending on its sign. This means that we can then say,

$$0\le a_n+\left|a_n\right|\le 2\left|a_n\right|$$

Now, since we are assuming that $\sum \left|a_n\right|$ is convergent then $\sum 2\left|a_n\right|$ is also convergent since we can just factor the 2 out of the series and 2 times a finite value will still be finite. This however allows us to use the Comparison Test to say that $\sum \left(a_n+\left|a_n\right|\right)$ is also a convergent series.

Finally, we can write,

$$\sum a_n=\sum \left(a_n+\left|a_n\right|\right)-\sum \left|a_n\right|$$

and so $\sum a_n$ is the difference of two convergent series and so is also convergent.

CONDITINALLY CONVERGENCE

More precisely, a series of real numbers $\sum _{�=0}^{\mathrm{\infty }}{�}_{�}$ is said to converge conditionally if $\underset{�\to \mathrm{\infty }}{lim}\sum _{�=0}^{�}{�}_{�}$ exists (as a finite real number, i.e. not ${\displaystyle \mathrm{\infty }}$ or ${\displaystyle -\mathrm{\infty }}$), but $\sum _{�=0}^{\mathrm{\infty }}\left|{�}_{�}\right|=\mathrm{\infty }.$

Example 1 Determine the radius of convergence and interval of convergence for the following power series.$$\sum _{n=1}^{\infty }\frac{{\left(x-6\right)}^n}{n^n}$$

SOLUTION

In this example the root test seems more appropriate. So,

$$\begin{array}{cc}L& =\underset{n\to \infty }{lim}{\left|\frac{{\left(x-6\right)}^n}{n^n}\right|}^{\frac{1}{n}}\\ & =\underset{n\to \infty }{lim}\left|\frac{x-6}{n}\right|\\ & =\left|x-6\right|\underset{n\to \infty }{lim}\frac{1}{n}\\ & =0\end{array}$$

So, since $L=0<1$ regardless of the value of $x$ this power series will converge for every $x$.

In these cases, we say that the radius of convergence is $R=\infty$ and interval of convergence is $-\infty <x<\infty$.

Example 2 Determine the radius of convergence and interval of convergence for the following power series.$$\sum _{n=1}^{\infty }\frac{2^n}{n}{\left(4x-8\right)}^n$$

SOLUTION

Let’s jump right into the ratio test.

$$\begin{array}{cc}L& =\underset{n\to \infty }{lim}\left|\frac{2^{n+1}{\left(4x-8\right)}^{n+1}}{n+1}\frac{n}{2^n{\left(4x-8\right)}^n}\right|\\ & =\underset{n\to \infty }{lim}\left|\frac{2n\left(4x-8\right)}{n+1}\right|\\ & =\left|4x-8\right|\underset{n\to \infty }{lim}\frac{2n}{n+1}\\ & =2\left|4x-8\right|\end{array}$$

So we will get the following convergence/divergence information from this.

$$\begin{array}{cc}& 2\left|4x-8\right|<1\text{series converges}\\ & 2\left|4x-8\right|>1\text{series diverges}\end{array}$$

We need to be careful here in determining the interval of convergence. The interval of convergence requires $\left|x-a\right|<R$ and $\left|x-a\right|>R$. 

$$\begin{array}{cc}& 8\left|x-2\right|<1\Rightarrow \left|x-2\right|<\frac{1}{8}\text{series converges}\\ & 8\left|x-2\right|>1\Rightarrow \left|x-2\right|>\frac{1}{8}\text{series diverges}\end{array}$$

So, the radius of convergence for this power series is $R=\frac{1}{8}$.

Now, let’s find the interval of convergence. Again, we’ll first solve the inequality that gives convergence above.

$$\begin{array}{c}-\frac{1}{8}<x-2<\frac{1}{8}\\ \frac{15}{8}<x<\frac{17}{8}\end{array}$$

Now check the endpoints.

$x=\frac{15}{8}$:
The series here is,

$$\begin{array}{cc}\sum _{n=1}^{\infty }\frac{2^n}{n}{\left(\frac{15}{2}-8\right)}^n& =\sum _{n=1}^{\infty }\frac{2^n}{n}{\left(-\frac{1}{2}\right)}^n\\ & =\sum _{n=1}^{\infty }\frac{2^n}{n}\frac{{\left(-1\right)}^n}{2^n}\\ & =\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^n}{n}\end{array}$$

This is the alternating harmonic series and we know that it converges.

$x=\frac{17}{8}$:
The series here is,

$$\begin{array}{cc}\sum _{n=1}^{\infty }\frac{2^n}{n}{\left(\frac{17}{2}-8\right)}^n& =\sum _{n=1}^{\infty }\frac{2^n}{n}{\left(\frac{1}{2}\right)}^n\\ & =\sum _{n=1}^{\infty }\frac{2^n}{n}\frac{1}{2^n}\\ & =\sum _{n=1}^{\infty }\frac{1}{n}\end{array}$$

This is the harmonic series and we know that it diverges.

So, the power series converges for one of the endpoints, but not the other. This will often happen so don’t get excited about it when it does. The interval of convergence for this power series is then,

$$\frac{15}{8}\le x<\frac{17}{8}$$