# Summation help

Tags: summation
 P: 837 I don't see how the following works: $$\sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n} = z^{-n_0}$$ I am missing the steps from $\sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n}$ to $z^{-n_0}$. If I try this step by step: $$\sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n} = \sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n_0} = z^{-n_0} \sum_{n=0}^\infty \delta ( n - n_0 )$$ Now, how is $\sum_{n=0}^\infty \delta ( n - n_0 )$ equal to 1. I don't get that. Thanks