Discrete Time Convolution Integral

Hello Everyone!

I've been stuck on this DT convolution sum for quite sum time, it's because I have not much experience in calculating DT sums. Well here goes:

$x[n] = (1/2)^{n}.u[n+2]$
$h[n] = (-3)^{n}.u[-n-1]$

Now, $x[n] \ast h[n] = \displaystyle \sum ^{\infty} _{k = -\infty} (1/2)^k .u[k+2] . (-3)^{n-k}.u[-n-1+k]$

Now here, I saw that whenevr we have step functions we can remove them and change the limits of the sum, so I modified the sum to be from n+1 to infinity as:
$\displaystyle \sum ^{\infty} _{k=n+1}(1/2)^k . (-3)^{n-k}$ .
But I think I am very wrong.

Any help would be greatly appreciated!