Question

In problems 1-6, determine the convergence set of the given power series.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{3}}{{\mathbf{n}}^{\mathbf{3}}}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{n}}$

Short answer

The set is,$x\in [1,3]$

Step-by-step solution

Step 1:To Find the Radius of convergence

Use the ratio test to determine the radius of convergence.

$\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{3}{n^3}}{\frac{3}{{(n+1)}^3}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{n^3}{n^3+3n^2+3n+1}\times \frac{3}{3}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{n^3}{n^3+3n^2+3n+1}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{1}{1+(3/n)+(3/n^2)+(1/n^3)}\right|\\ =1\end{array}$

The radius of convergence is 1, therefore convergent set for the given power series is.$|x-2|<1$

Find the set of convergence

To completely identify the convergence set, we have to check whether the boundary points 1 and 3 are included in the set or not.

Checking at ,$x=1$by substituting the value of x by 1,

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{3}{n^3}{\left(x-2\right)}^n=\sum _{n=0}^{\infty }\frac{3}{n^3}{\left(1-2\right)}^n\\ =\sum _{n=0}^{\infty }\frac{3}{n^3}{\left(-1\right)}^n\end{array}$

The above series is an alternating power series with$p=3>1$, which is convergent, thus the point 1 is included in the convergent set.

Similarly, checking at$x=3$, by substituting the value of x by 3,

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{3}{n^3}{\left(x-2\right)}^n=\sum _{n=0}^{\infty }\frac{3}{n^3}{\left(3-2\right)}^n\\ =\sum _{n=0}^{\infty }\frac{3}{n^3}\end{array}$

The above series is a power series with$p=3>1$, which is convergent, thus the point 3 is included in the convergent set.

The convergent set for the given power series is.$x\in [1,3]$