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Let $$S_m=\sum _{n=1}^m\frac{n^2+n-1}{(n+2)!}.$$We compute the first few sums $S_m$: \renewcommand{\arraystretch}{1.5} \begin{array}{c|c} m & S_m \\ \hline 1 & \frac{1}{6} \\ 2 & \frac{3}{8} \\ 3 & \frac{7}{15} \\ 4 & \frac{71}{144} \\ 5 & \frac{419}{840} \end{array}We note that the fractions seem to be approaching $\frac{1}{2},$ so we also compute $\frac{1}{2}-S_m$: \renewcommand{\arraystretch}{1.5} \begin{array}{c|c|c} m & S_m & \frac{1}{2} - S_m \\ \hline 1 & \frac{1}{6} & \frac{1}{3} \\ 2 & \frac{3}{8} & \frac{1}{8} \\ 3 & \frac{7}{15} & \frac{1}{30} \\ 4 & \frac{71}{144} & \frac{1}{144} \\ 5 & \frac{419}{840} & \frac{1}{840} \end{array}We can relate the fractions $\frac{1}{2}-S_m$ to factorials in the following way: $$\frac{1}{3}=\frac{2}{3!},\frac{1}{8}=\frac{3}{4!},\frac{1}{30}=\frac{4}{5!},\frac{1}{144}=\frac{5}{6!},\frac{1}{840}=\frac{6}{7!}.$$Thus, we conjecture that $$S_m=\frac{1}{2}-\frac{m+1}{(m+2)!}.$$So, let $$T_n=\frac{1}{2}-\frac{n+1}{(n+2)!}.$$Then $$\begin{array}{rl}{T}_n-T_{n-1}& =\left(\frac{1}{2}-\frac{n+1}{(n+2)!}\right)-\left(\frac{1}{2}-\frac{n}{(n+1)!}\right)\\ & =\frac{n}{(n+1)!}-\frac{n+1}{(n+2)!}\\ & =\frac{n(n+2)-(n+1)}{(n+2)!}\\ & =\frac{n^2+n-1}{(n+2)!},\end{array}$$which is exactly what we are summing.

From the identity $$\frac{n}{(n+1)!}-\frac{n+1}{(n+2)!}=\frac{n^2+n-1}{(n+2)!},$$we have that $$\begin{array}{rl}\sum _{n=1}^{\infty }\frac{n^2+n-1}{(n+2)!}& =\left(\frac{1}{2!}-\frac{2}{3!}\right)+\left(\frac{2}{3!}-\frac{3}{4!}\right)+\left(\frac{3}{4!}-\frac{4}{5!}\right)+\cdots \\ & =\boxed{\frac{1}{2}}.\end{array}$$