91Ó°ÊÓ

Use the ratio test to determine the radius of convergence.

$\begin{array}{c}\underset{n→\mathrm{∞}}{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n→\mathrm{∞}}{\lim }\left|\frac{\frac{3}{n^3}}{\frac{3}{{(n+1)}^3}}\right|\\ =\underset{n→\mathrm{∞}}{\lim }\left|\frac{n^3}{n^3+3n^2+3n+1}×\frac{3}{3}\right|\\ =\underset{n→\mathrm{∞}}{\lim }\left|\frac{n^3}{n^3+3n^2+3n+1}\right|\\ =\underset{n→\mathrm{∞}}{\lim }\left|\frac{1}{1+(3/n)+(3/n^2)+(1/n^3)}\right|\\ =1\end{array}$

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

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,

$\begin{array}{c}\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{3}{n^3}{\left(x-2\right)}^n=\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{3}{n^3}{\left(1-2\right)}^n\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{3}{n^3}{\left(-1\right)}^n\end{array}$

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,

$\begin{array}{c}\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{3}{n^3}{\left(x-2\right)}^n=\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{3}{n^3}{\left(3-2\right)}^n\\ =\underset{n=0}{\overset{\mathrm{∞}}{∑}}\frac{3}{n^3}\end{array}$

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.$x∈[1,3]$