1.05^{n}=2

nlog1.05=log2

n=log2/log1.05=14.20669908

I did the second part separately before. Here it is

Assuming that x^{2}-x-1≠1 the finite sum G=Σ_{k=0}^{k=N-1}(x^{2}-x-1)k=(1-(x^{2}-x-1)^{N})/(1-(x^{2}-x-1)) .

The only way that this sum converges as N→∞ is for (x^{2}-x-1)^{N}→0 or that |x^{2}-x-1|<1

To force this means that x^{2}-x-1<1 and x^{2}-x-1>-1

x^{2}-x-1<1 means that x^{2}-x-2<0. That is between the roots of this parabola. x^{2}-x-2=(x+1)(x-2)=0 between

x=-1 or x=2. So on the interval -1<x<2.; For the second condition x^{2}-x-1>-1, which is x^{2}-x>0, that is outside of the interval between the the roots of this parabola. x^{2}-x=x(x-1)=0 when x=0 or x=1. So on the intervals

(-∞,0) or (1,∞)

The intersection of these intervals is (-1<x<0)∪(1<x<2). For x in this set the infinite sum converges.