\begin{aligned}
 \sum_{i\in\omega} (i+2)^{\omega+i}
 & = \sup_{n\in\omega} \sum_{i\in n} \omega\cdot (i+2)^i \\
 & = \sup_{n\in\omega} (\omega\cdot a_n) \\
 & = \omega^2
\end{aligned}