Let's progress Geometrically!

We can calculate S and Sn as sums of a geometric progression (to infinity and to n, respectively):

$S = \frac {1}{1- \frac{1}{2} } = 2$

$S_n = \frac { 1 - ( \frac {1}{2} ) ^n }{1- \frac{1}{2} } = 2 - ( \frac {1}{2} ) ^{n-1}$

$S - S_n < \frac {1}{1000}$

$2 - (2 - ( \frac {1}{2} ) ^{n-1}) < \frac {1}{1000}$

$( \frac {1}{2} ) ^{n-1} < \frac {1}{1000}$

$1000 < 2^{n-1}$

$log_2(1000) < n - 1$

$9.965784 < n -1$

$10.965784 < n$

$\text {Hence, our answer should be: } \boxed {11}$

$\begin{aligned} S = \frac{1}{1-\frac{1}{2}} &\implies \boxed{2} \\ \\ S_{n} = \frac{1 - (1/2)^n}{1-1/2} &\implies \boxed{2- \bigg(\frac{1}{2}\bigg)^{n-1}} \end{aligned}$

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$\begin{aligned} S - S_{n} & < \frac{1}{1000} \\ S - S_{n} &= 2 - \bigg(2- \bigg(\frac{1}{2}\bigg)^{n-1}\bigg) \\ \bigg(\frac{1}{2}\bigg)^{n-1} & < \frac{1}{1000} \\ \frac{1}{2^n} &< \frac{1}{2000} \\ 2^n &> 2000 \\ n &> 10.96 \\ n &= \boxed{11} \end{aligned}$

S = 2, S n = [(2^n) -1]/(2^(n - 1))), S - S n = 1/(2^(n - 1)). If this is less than 1/1000, then 2^(n - 1) = 1024, n - 1 = 10, n = 11.