## Summation help

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

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 Recognitions: Homework Help Science Advisor delta(n-n0) is equal to 1 if n=n0 and zero otherwise. So the only way the sum could be nonzero is if n0 is a positive integer. Is n0 a positive integer?

 Quote by Dick delta(n-n0) is equal to 1 if n=n0 and zero otherwise. So the only way the sum could be nonzero is if n0 is a positive integer. Is n0 a positive integer?
:) - wow, i've been looking at this crap for too long. I can't believe I missed that.

Thanks man :) Yeah, n0 is a positive integer.

time for a break...