Convolution computing How we can compute the convolution product $$\Big(\sum_{n=0}^{+\infty} \delta_n^{(n)}\Big) \star \Big(\sum_{n=0}^{+\infty} \delta_n\Big)$$
where $\delta$ is Dirac distribution?
Thank's for the help
 A: Formal computation. For every test function, $\varphi\star \delta_n $ is the shifted function $t\mapsto \varphi(t-n) $; the shift is $n$ units to the right.  Let's  accept that the same holds for distribution. Then 
$$\delta_k \star\left(\sum_{n=0}^{+\infty} \delta_n^{(n)}\right) = \sum_{n=0}^{+\infty} \delta_{n+k}^{(n)}$$
And formal summation over $k$ yields
$$\left(\sum_{k=0}^\infty \delta_k \right)\star \left(\sum_{n=0}^{+\infty} \delta_n^{(n)} \right) = \sum_{k=0}^\infty \sum_{n=0}^{+\infty} \delta_{n+k}^{(n)}$$
The end result can be rewritten in a way that shows we indeed have a distribution, by letting $m=n+k$: 
$$\sum_{m=0}^\infty \sum_{n=0}^{m} \delta_{m}^{(n)} \tag{1}$$
The main point here is that on every compact subset only finitely many terms of (1) are present. 

If you want to make the above rigorous, take a (compactly supported) test function $\varphi$, evaluate (1) against it (it's a finite sum). On the other hand, the evaluation of convolution $f\star g$   amounts to applying the product distribution $f(x)g(y)$ to the function $f(x+y)$ in the plane. This gives the same result.

As an aside: if the convolution was instead 
$$\Big(\sum_{n=0}^{+\infty} \delta_n^{(n)}\Big) \star \Big(\sum_{n=0}^{+\infty} \delta_{-n}\Big)$$
the formula would not meaningul (at least I don't see a way to define this convolution meaningfully).
