Let \(b_k = \frac{k^2+1}{k^4+2k+2}\) and \(a_k = \frac{1}{k^2}\text{.}\) Then
\begin{align*}
\lim_{k \to \infty} \frac{b_k}{a_k} \amp = \lim_{k \to \infty} \frac{\frac{k^2+1}{k^4+2k+2}}{\frac{1}{k^2}}\\
\amp = \lim_{k \to \infty} \frac{k^4+k^2}{k^4+2k+2}\\
\amp = \lim_{k \to \infty} \frac{1+\frac{1}{k^2}}{1+\frac{2}{k^3}+\frac{2}{k^4}}\\
\amp = 1\text{.}
\end{align*}
Since \(\lim_{k \to \infty} \frac{b_k}{a_k}\) is a finite positive constant, the Limit Comparison Test shows that \(\sum \frac{1}{k^2}\) and \(\sum \frac{k^2+1}{k^4+2k+2}\) either both converge or both diverge. We know that \(\sum \frac{1}{k^2}\) is a \(p\)-series with \(p > 1\) and so \(\sum \frac{1}{k^2}\) converges. Therefore, the series \(\sum \frac{k^2+1}{k^4+2k+2}\) also converges.