The McLaurin series is just the Taylor expansion of $F(x)$ about $x=0$
If you're clever you'll notice $F(x)$ is the sum of a well known series.
Note that F(1) is undefined and indeed the series will not converge for x=1.
Well I haven't seen Taylor series before only learnt about Mclaurin series, ive got little knowledge as I havnt learnt this before, would help if you could explain it, this is just for reference as the question ive got to solve is x=0.01
I don't understand how you can be given a problem without having been taught the reference material.
The Mclaurin series way of doing this problem is to recognize that
$\displaystyle{\sum_{k=0}^\infty}x^n = \dfrac 1 {1-x}~~\mbox{ for } |x| < 1$
This is the geometric series. Have you seen this before?
So given that your $F(x)=\dfrac 1 {1-x}$ it's obvious that your Mclaurin series is
$F(x) = \displaystyle{\sum_{k=0}^\infty}x^n$
and this converges for $|x| < 1$
It's not possible to try to evaluate this at $x=0.01$ via the series as there an infinite number of terms. So just plug $0.01$ into the formula for the sum, i.e. $F(x)$.
I'll let you compute $\dfrac 1 {1-0.01}$