There is a representation of the delta function as the limit:
$$ \delta(x) = \lim_{\epsilon\rightarrow 0} \left [ \int_{-\infty}^\infty dx \exp \left (- \frac{x^2}{2 \epsilon} \right ) \right ]^{-1} \exp \left (- \frac{x^2}{2 \epsilon} \right ) = \lim_{\epsilon\rightarrow 0} \frac{1}{\sqrt{2\pi\epsilon}} \exp \left (- \frac{x^2}{2 \epsilon} \right ) $$
Is the following also a valid "discrete" representation?
$$ \delta(n) = \lim_{\epsilon\rightarrow 0} \left [ \sum_{n=-\infty}^\infty \exp \left (- \frac{n^2}{2 \epsilon} \right ) \right ]^{-1} \exp \left (- \frac{n^2}{2 \epsilon} \right ) $$
If you mean what I think you mean, then the answer is yes. In contrast to the continuous case, here the functions converge to a function, namely $\delta(n)=\delta_{n0}$ (the Kronecker symbol). If this is the function that you wanted to "represent", then you did. I'd expect the usefulness of such a "representation" to be more limited than in the continuous case, where a distribution is represented by a sequence of functions.