Impulse Response to LTI
There is a general question and a specific question.

To answer the specific question, you can guess at an LTI and check if it satisfies your spec:

$$ h[n] = a^n u[n]  \qquad \text{where} \ |a| &lt; 1 $$

Here is my guess:

$$ y[n] = a \cdot y[n-1] + x[n]  \qquad \text{where} \ y[-1]=0 $$

you can set $x[n] = \delta[n]$ and you will find that $y[n] = h[n]$ for all integer $n$ and you are done with the specific question.

To see if a conceptual discrete-time LTI can be made to satisfy a general impulse response spec, $h[n]$, there are a few requisite properties to satisfy:


Is the system causal?  This means the output $y[n]$ does not react to future samples of the input $x[n]$.  And if it's both LTI and causal, that means the impulse response does not react to the input impulse before that input impulse takes on non-zero values.  And that means $$ h[n] = 0 \qquad \forall \ n&lt;0$$  You can see that this is satisfied by your example spec.
Is the LTI system FIR or IIR?  If it's FIR, that means that the non-zero length of the impulse response, $h[n]$ is finite in length (let's call that FIR length "$M$") and that means $$ h[n] = 0 \qquad \forall \ n&lt;0, \ n \ge M$$ Now it turns out that your example is not FIR (which means it is IIR, and Infinite Impulse Response LTI system), but if it were FIR, then your general system can always be realized (or "actualized") as $$ y[n]=\sum\limits_{m=0}^{M-1} h[m] \, x[n-m] $$  That can always be done for an FIR.  It might not be the most efficient way to make an FIR for extremely large $M$, but it is a method that, if you can afford it, will always work.
If it's an IIR (which means the condition in paragraph 2 above is not satisfied), then you have some restrictions about what can be realized and what cannot.  And you have some other conditions to nail down.  One is "BIBO stability" which means, for any bounded input $x[n]$, the output $y[n]$ is also bounded.  Now, I don't know all of the implications, but that does mean that systems that blow up, like $$ h[n] = a^n u[n]  \qquad \text{where} \ +1 &lt; |a| $$ can be realized (same input/output equation as in 2), but they will blow up.  I don't think you want that.  BIBO stability also means that impulse responses like $$ h[n] = \tfrac{1}{n+1} u[n] $$ will also blow up because $\tfrac{1}{n+1}$ does not converge fast enough.  If, say, $x[n]=u[n]$ (which is bounded), the output $y[n]$ will blow up.
So it turns out that BIBO stable LTI will need to either be FIR (they're always stable from an input/output POV) or, if the LTI is IIR, then the impulse response must decay exponentially.  That means, for large $n$, $$  \Big|h[n] \Big| \le A \ a^n \ u[n] \qquad \text{for some} \ |a| &lt; 1  $$  If that's true, you can conceptually make an IIR filter that approximates any $h[n]$ that satisfies the causality condition (paragraph 1) and the BIBO stability conditions (paragraph 3).  The way we learn to do that is by taking a DSP course after learning LTI system theory.