This question comes from Strikwerda, "Finite Difference Schemes and Partial Differential Equations":
By multiplying $$ \frac{v^{n+1}_m - v^{n}_m}{k} + \frac{a}{2}\left( \frac{v^{n+1}_{m+1}-v_{m}^{n+1}}{h} + \frac{v^{n}_{m+1}-v_{m-1}^{n}}{h} \right) = 0 $$ by $v^{n+1}_m + v^n_m$ and summing over all $m$ conclude the scheme is stable for $a\lambda <1$, where $\lambda = \frac{k}{h}$.
I have worked this out to the following relation: $$ \sum_{m=-\infty}^\infty \left(1 - \frac{a\lambda}{2} \right)(v^{n+1}_m)^2 + \frac{a\lambda}{2}v^{n+1}_m v^{n+1}_{m+1} = \sum_{m=-\infty}^\infty \left(1 - \frac{a\lambda}{2} \right)(v^{n}_m)^2 + \frac{a\lambda}{2}v^{n}_m v^{n}_{m+1} $$ but now I have difficulty getting an expression bounding the righthand side of the form $\sum |v^n_m|^2 $ due to the cross-terms of $m$ and $m+1$.
My suspicion is that the final form will be something along the lines of $f(a\lambda)^n\sum|v^0_m|^2$, where $v^0_m$ is the discrete initial condition and $f$ is some algebraic expression, so as $n\to \infty$ we can derive the restriction on $a\lambda$ from $f(a\lambda)<1$.
A note that this exercise is before von Neumann analysis is introduced, so it this should be possible to show through inequalities and algebraic manipulation.