Consider the following sum:
$$\sum_{j=1}^{\infty} \frac{\theta^j}{j!} \cdot e^{-\theta} \cdot j$$
This should reduce to $\theta$.
My approach:
$$= e^{-\theta} \cdot \sum_{j=1}^{\infty} \frac{\theta^j}{j!} \cdot j$$ $$= e^{-\theta} \cdot \sum_{j=1}^{\infty} \frac{\theta^j}{(j-1)!}$$ $$= e^{-\theta} \cdot \theta \cdot \sum_{j=1}^{\infty} \frac{\theta^{(j-1)}}{(j-1)!}$$ $$= e^{-\theta} \cdot \theta \cdot e^{\theta}$$ $$= \theta$$
I think there is a shortcut involved, i.e. in the second last step I am "cheating" a bit. Would I not need $\sum_{j=1}^{\infty} \frac{\theta^{j}}{j!}$ in order to plug in $e^{\theta}$ for the taylor series?