Answer
\[\text{The series converges}\]
Work Step by Step
\[\begin{align}
& \sum\limits_{k=3}^{\infty }{\frac{1}{{{\left( k-2 \right)}^{4}}}}=\frac{1}{{{1}^{4}}}+\frac{1}{{{2}^{4}}}+\frac{1}{{{3}^{4}}}+\frac{1}{{{4}^{4}}}+\cdots \\
& \text{We can write the series as} \\
& \sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{4}}}} \\
& \text{Using the p-series test }\left( \text{Theorem 8}\text{.11}\text{, page 632} \right) \\
& \text{The }p\text{-series }\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{p}}}\text{ converges for }p>1\text{ and diverges for }p\le 1} \\
& \text{Comparing} \\
& \underbrace{\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{4}}}}}_{\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{p}}}}}\Rightarrow p=4 \\
& p>1,\text{ then the series converges} \\
\end{align}\]
