Proof that $ y(n) = ∑_{k=-∞}^{∞}\ {a}^{-k}u(n-k)u(-k) = \frac{1}{1-a }$ if $n>0$

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Can someone explain the steps and how the boundaries for the summation change to result in the answer (And possible for the case where $n\leq 0$. I am not really a mathematician, don't know if the question is clear( this question comes from finding the impulse response of an LTI system )

The full equation is

$$ y(n) = \sum_{k=-\infty}^{\infty} {a}^{-k}u(-k)u(n-k) = \frac{1}{1-a} $$

if $n>0$ where $u(n)$ is the unit step function.

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The product of the step-functions (assuming $u$ is the Heaviside step-function) $u(-k)u(n-k)$ is non-zero only if $-k\geq 0$ and $n-k\geq 0$, that is if $n\geq k$ and $k\leq 0$. If $n > 0$ then this is the case for all negative $k$ so the sum becomes (assuming $|a|<1$)

$$\sum_{k=-\infty}^\infty a^{-k}u(-k)u(n-k) = \sum_{k=-\infty}^0 a^{-k} = \sum_{k=0}^\infty a^{k} = \frac{1}{1-a} $$

where in the second to last equality I switched the summation index $k\to -k$ and in the last equality I used that the sum of a geometric series $1+x+x^2+\ldots = \frac{1}{1-x}$ if $|x|<1$.