The given series is
\(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}\)

The series \(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}\) can be simplified as follows.

\(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}=\sum_{k=3}^\infty\frac{2}{\frac{k^2-k-2}{k^2}}\)

\(\sum_{k=3}^\infty\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

\(\sum_{k=3}^\infty a_k\) where \(a_k=f(k)\)

Consider the function \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

Here \(\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\ne0\) for all values of k in the interval \([3,\infty)\), so this function is continuous of k

Put 3 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

\(f(3)=\frac{2}{1-\frac{1}{3}-\frac{2}{3^2}}\)

\(f(3)=\frac{2}{1-\frac{1}{3}-\frac{2}{9}}\)

\(=\frac92\)

Put 4 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

\(f(4)=\frac{2}{1-\frac{1}{4}-\frac{2}{16}}\)

\(=\frac{16}{5}\)

Put 5 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

\(f(5)=\frac{2}{1-\frac{1}{5}-\frac{2}{5^2}}\)

\(f(5)=\frac{2}{1-\frac{1}{5}-\frac{2}{25}}\)

\(=\frac{25}{9}\)

It is observed that the function is a deceasing function,

Therefore, it can be concluded that the Integral test is used to determine the convergence of the given series.

The series \(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}\) can be simplified as follows.

\(\sum_{k=3}^\infty\frac{2k^2}{k^2-k-2}=\sum_{k=3}^\infty\frac{2}{\frac{k^2-k-2}{k^2}}\)

\(\sum_{k=3}^\infty\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

\(\sum_{k=3}^\infty a_k\) where \(a_k=f(k)\)

Consider the function \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

Here \(\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\ne0\) for all values of k in the interval \([3,\infty)\), so this function is continuous of k

Put 3 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

\(f(3)=\frac{2}{1-\frac{1}{3}-\frac{2}{3^2}}\)

\(f(3)=\frac{2}{1-\frac{1}{3}-\frac{2}{9}}\)

\(=\frac92\)

Put 4 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

\(f(4)=\frac{2}{1-\frac{1}{4}-\frac{2}{16}}\)

\(=\frac{16}{5}\)

Put 5 for k in \(f(k)=\frac{2}{1-\frac{1}{k}-\frac{2}{k^2}}\)

\(f(5)=\frac{2}{1-\frac{1}{5}-\frac{2}{5^2}}\)

\(f(5)=\frac{2}{1-\frac{1}{5}-\frac{2}{25}}\)

\(=\frac{25}{9}\)

It is observed that the function is a deceasing function,

Therefore, it can be concluded that the Integral test is used to determine the convergence of the given series.