Question

If $S_n=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.........+\frac{1}{2^{n-1}}$; Calculate the least value of $n$ such that $2-S_n<\frac{1}{100}$

Answer 1

We can see that the terms are in G.P with first term $1$ and the common ratio $\frac{1}{2}$
Hence $S_n=\frac{{\left(\frac{1}{2}\right)}^n-1}{\frac{1}{2}-1}=2\frac{\{2^n-1\}}{2^n}=2-\frac{1}{2^{n-1}}\Rightarrow 2-S_n=\frac{1}{2^{n-1}}$
Hence A/Q
$\frac{1}{2^{n-1}}<\frac{1}{100}\Rightarrow 2^{n-1}>100\Rightarrow n-1>6\Rightarrow n>7$
Therefore, $n=8$