Stability range for h[n] = e^(an)u[n] + e^(bn)u[−n]

An LTI system is BIBO stable if and only if its impulse response is absolutely summable:

$$S=\sum_{n=-\infty}^{\infty}|h[n]|<\infty$$

Here $h[n]=e^{an}u[n]+e^{bn}u[-n]$ (take $a,b$ real). Split the sum:

$$\begin{aligned} S &= \sum_{n=0}^{\infty} e^{an} + \sum_{n=-\infty}^{0} e^{bn} \\ &= \sum_{n=0}^{\infty} (e^{a})^{n} + \sum_{m=0}^{\infty} (e^{-b})^{m} \qquad (m=-n) \end{aligned}$$

(At $n=0$ both terms are present; that adds only a finite value 2, so it does not affect convergence.)

Right-sided (causal) part

A geometric series $\sum r^n$ converges only if $|r|<1$:

$$e^{a}<1 \;\Rightarrow\; a<0$$

Its sum is then $\frac{1}{1-e^{a}}$.

Left-sided (anti-causal) part

$$e^{-b}<1 \;\Rightarrow\; -b<0 \;\Rightarrow\; b>0$$

Its sum is then $\frac{1}{1-e^{-b}}$.

Result

Both series must converge, so

$$S=\frac{1}{1-e^{a}}+\frac{1}{1-e^{-b}}<\infty \quad \text{only when } a<0 \text{ and } b>0$$

Answer: the system is stable for $a<0$ and $b>0$ (if $a,b$ are complex: $\mathrm{Re}(a)<0$ and $\mathrm{Re}(b)>0$). Physically, the part for $n\ge0$ must decay as $n\to+\infty$ and the part for $n\le0$ must decay as $n\to-\infty$. Note that the system is non-causal because $h[n]\ne0$ for $n<0$.