Answer
The interval of convergence is $\left[ {2,4} \right]$.
Work Step by Step
Let ${a_n} = \dfrac{{{n^6}}}{{{n^8} + 1}}{\left( {x - 3} \right)^n}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{{\left( {n + 1} \right)}^6}}}{{{{\left( {n + 1} \right)}^8} + 1}}\dfrac{{{{\left( {x - 3} \right)}^{n + 1}}}}{{{{\left( {x - 3} \right)}^n}}}\dfrac{{{n^8} + 1}}{{{n^6}}}} \right|$
$ = \left| {x - 3} \right|\mathop {\lim }\limits_{n \to \infty } \left| {{{\left( {\dfrac{{n + 1}}{n}} \right)}^6}\dfrac{{\left( {{n^8} + 1} \right)}}{{\left( {{{\left( {n + 1} \right)}^8} + 1} \right)}}} \right|$
$ = \left| {x - 3} \right|\mathop {\lim }\limits_{n \to \infty } \left| {{{\left( {1 + \dfrac{1}{n}} \right)}^6}\dfrac{{\left( {1 + \dfrac{1}{{{n^8}}}} \right)}}{{\left( {{{\left( {1 + \dfrac{1}{n}} \right)}^8} + \dfrac{1}{{{n^8}}}} \right)}}} \right|$
$ = \left| {x - 3} \right|$
The series converges if $\rho = \left| {x - 3} \right| \lt 1$. That is,
$ - 1 \lt x - 3 \lt 1$
or $2 \lt x \lt 4$.
Next, we check the endpoints: $x=2$ and $x=4$.
1. $x=2$: ${\ \ \ \ \ }$ $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{n^6}}}{{{n^8} + 1}}{\left( {x - 3} \right)^n} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}\dfrac{{{n^6}}}{{{n^8} + 1}}$
Since the sequence $\left\{ {\dfrac{{{n^6}}}{{{n^8} + 1}}} \right\}$ is positive, decreasing and converge to $0$, by the Alternating Series Test, this series converges.
2. $x=4$: ${\ \ \ \ \ }$ $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{n^6}}}{{{n^8} + 1}}{\left( {4 - 3} \right)^n} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{n^6}}}{{{n^8} + 1}}$
We have the inequality: $\dfrac{{{n^6}}}{{{n^8} + 1}} \lt \dfrac{{{n^6}}}{{{n^8}}}$.
Or $\dfrac{{{n^6}}}{{{n^8} + 1}} \lt \dfrac{1}{{{n^2}}}$.
The larger series $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{n^2}}}$ converges (because $p = 2 \gt 1$ by the Convergence of $p$-Series).
Therefore, by the Direct Comparison Test, the smaller series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{{n^6}}}{{{n^8} + 1}}$ also converges.
Thus, the interval of convergence is $\left[ {2,4} \right]$.
