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

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{4}}{{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{n}}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{)}}^{\mathbf{n}}$

Use the ratio test to determine the radius of convergence.

$\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{4}{n^2+2n}}{\frac{4}{{(n+1)}^2+2(n+1)}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{n^2+2n}{n^2+2n+1+2n+2}\times \frac{4}{4}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{n^2+2n}{n^2+4n+3}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{n^2(1+(2/n\left)\right)}{n^2\left(1+(4/n)+(3/n^2)\right)}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{(1+(2/n\left)\right)}{\left(1+(4/n)+(3/n^2)\right)}\right|\\ =1\end{array}$

The radius of convergence is 1, therefore convergent set for the given power series is .

$|x-3|<1$

To completely identify the convergence set, we have to check whether the boundary points 2 and 4 are included in the set or not.

Checking at $x=2$, by substituting x by 2,

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{4}{n^2+2n}{\left(x-3\right)}^n=\sum _{n=0}^{\infty }\frac{4}{n^2+2n}{\left(2-3\right)}^n\\ =\sum _{n=0}^{\infty }\frac{4}{n^2+2n}{\left(-1\right)}^n\end{array}$

The above series is an alternating harmonic series, which is convergent in nature, thus the point 2 is included in the convergent set.

Similarly, checking at x-4, by substituting x by 4 ,

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{4}{n^2+2n}{\left(x-3\right)}^n=\sum _{n=0}^{\infty }\frac{4}{n^2+2n}{\left(4-3\right)}^r\\ =\sum _{n=0}^{\infty }\frac{4}{n^2+2n}\end{array}$

Since, $n^2+2n>n^2\Rightarrow \frac{1}{n^2+2n}<\frac{1}{n^2}$and$\sum \frac{1}{n^2}$is convergent, then, by the comparison test it follows that$\sum \frac{1}{n^2+2n}$ is also convergent.

The convergent set for the given power series is.$x\in [2,4]$