How do I prove
$w_{k+1} = \frac{-(2 (3 - 4 e + 2 e^2) g )}{(-5 + 4 e)} \times w_k$
converges to $0$ over $k\rightarrow \infty$ and say $w_0=1$
Possible bounds are: $0<e<1$ and $0<g<1$.
Also $e$ is just another variable, not Euler number
How do I prove
$w_{k+1} = \frac{-(2 (3 - 4 e + 2 e^2) g )}{(-5 + 4 e)} \times w_k$
converges to $0$ over $k\rightarrow \infty$ and say $w_0=1$
Possible bounds are: $0<e<1$ and $0<g<1$.
Also $e$ is just another variable, not Euler number
For $g=0.9$ and $e=0.1$, then $$\left| \frac{-(2 (3 - 4 e + 2 e^2) g )}{(-5 + 4 e)}\right| >1$$
So the sequence diverges.