smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$ 
If $S_n$ and $S_\infty$ are sums to $n$ terms and sum to infinity of a geometric progression $3,-\frac{3}{2},\frac{3}{4},...$ respectively, find the smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$

My attempt,
$$|S_n-S_\infty|<0.001$$
$$|\frac{3[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}-\frac{3}{1-(-\frac{1}{2})}|<0.001$$
$$|2[1-(-\frac{1}2)^n]-2|<0.001$$
How to proceed? Thanks in advance.
 A: $$|2[1-(-\frac{1}2)^n]-2|<0.001\\
|-2(\frac{-1}2)^n|<0.001\\
+2|(\frac{-1}2)^n|<0.001\\\to \text{abs function properties  } |\frac{-1}{2^n}|=\frac{1}{2^n}\\
+2.\frac{1}{2^n}<\frac{1}{1000}\\
\frac{2^n}{2}>1000\\
2^{n-1}>1000\\\text{note that  } 2^{10}=\color{red} {1024>1000}\\\to \\n-1\geq 10\\n \geq 11$$
A: $$|\frac{3[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}-\frac{3}{1-(-\frac{1}{2})}|<0.001$$
$$|2[1-(-\frac{1}2)^n]-2| = |2(-\frac{1}2)^n|=\frac{1}{2^{n-1}}<0.001$$
ie 
$$-(n-1) \ln 2  < \ln (0.001) \implies n-1> \frac{\ln 1000 }{\ln 2} $$
hence the smalest integer is given by 
$$N=\left[\frac{\ln 1000 }{\ln 2} +1 \right] +1 = 11$$ 
$[x]$ stand for the floor of $x$.
A: For a convergent alternating sum $S_n=\sum\limits_{k=1}^na_k$, with $S=\lim\limits_{n\to\infty}S_n$, we have
$$\begin{align*}
|S_n-S|&=\left|\sum_{k=1}^na_k-S\right|\\
&=\left|\sum_{k=n+1}^\infty a_k\right|\\
&\le\left|\sum_{k=n+1}^Na_k\right|&\text{for some }N>n+1\\
&=|a_{n+1}+\cdots+a_N|\\
&\le|a_{n+1}|+\cdots+|a_N|\\
&\le|a_{n+1}|
\end{align*}$$
We have $a_k=3\left(-\frac12\right)^{k-1}$ for $k\in\mathbb N$. Thus
$$|S_n-S|\le\left|3\left(-\frac12\right)^n\right|=\frac3{2^n}$$
Then the approximate value of $S_n$ will fall within $0.001$ of $S$ for
$$\frac3{2^n}<0.001\implies2^n>3000$$
$2^{11}=2048$ and $2^{12}=4096$, so $n=12$ will guarantee $|S_n-S|<0.001$.
Unfortunately, as the other answers demonstrated, this isn't a hard limit. So perhaps this isn't the best method, but it gives a good starting point to find a better $n$.
