I assume we have a function f(n)=mf(n-1) or something like it, where -1<m<0.

For example, if m=-½, and f₁=1, f₂=-½f(1)=-½.

f₃=-½f₂=¼, f₄=-⅛ etc.

If f(1)=a, in the general case, f₂=-a/2, f₃=a/4, f₄=-a/8, so f(n)=(-1)ⁿ⁻¹a/2ⁿ⁻¹.

As n gets bigger, f(n)→0 no matter what the initial value of a was, provided a≠0.

For a general linear function f(n)=mf(n-1)+b where b is any non-zero number and f(1)=a≠0, we have:

f₂=ma+b, f₃=m(ma+b)+b=m²a+mb+b=m²a+b(1+m), f₄=m³a+b(1+m+m²), ...

f(n)=mⁿ⁻¹a+b(1+m+m²+...+mⁿ⁻²)=mⁿ⁻¹a+b(mⁿ⁻¹-1)/(m-1).

So, since -1<m<0, the first term converges to 0 but the second term converges to a finite non-zero value for b≠0, because we can write this as b/(1-m). The denominator is positive because m is negative and mⁿ⁻¹→0. If b is positive the convergence is to a positive value, otherwise it’s negative.