What are the Convolution Intervals for Step Function and Exponential Decay?

[MalB]

I am trying to see if I understand the convolution process correctly.
This is from the solved example of my prof's notes:

x[n] = u[n]
and h[n] = a^-nu[-n] for 0<a<1

As expected the first step was

y[n] = x[n]* h[n]

= Ʃ h[k]x[n-k] -∞<k<∞
= Ʃ a^-ku[-k]u[n-k] -∞<k<∞

From my understanding u[-k] = 0 for all k > 0 so that would give us the upper limit of k to be 0
And we would also need for n-k > 0 and when we have those conditions we would just then be left with the summation of a^-k since the other 2 would always evaluate to 1 in that interval.

But for the solution the cases considered were when n <= 0 and when n >0 and then the answers were

Ʃ a^k -n<k<∞ for the first case (n<=0)
and Ʃa^k 0<k<∞

I don't understand how they came to those intervals. Any help would be appreciated.

 

[MalB]

I think i have it figured out.
I will post my understanding and then maybe someone can comment on whether I was right.
My idea of k<0 was right but only if n > 0 which would make the second term u[n-k] always positive also giving us a SUM a^-k and then we flip the summation around cause we change the sign and end up with the answer noted for n>0

for n<0 for n-k >= 0 we need -k >= n or k<= -n
and then we end up with SUM a^-k and then again flip the limits and end up with the answer noted for n<0