Find sum of  $n$ term, sum to infinity,  least value for $n$

The $n$th  term in geometric series is given by
\[a_n=aq^{n-1},   n\geq 1.\]
Since the first term is 20, we have $a=20$. We also have
\[-\frac{5}{2}=a_4=20q^3   \Rightarrow q^3=-\frac{5}{40}=-\frac{1}{8}  \Rightarrow q=-\frac{1}{2}.\]
Thus
\[a_n=20(-\frac{1}{2})^{n-1}.\]
(i) The sum of the first $n$ terms is
\[S_n=\sum_{i=1}^{n}20(-\frac{1}{2})^{n-1}=\frac{20(1- (-\frac{1}{2})^n )}{1- (-\frac{1}{2})}=\frac{40}{3}(1- (-\frac{1}{2})^n).\]
(ii) 

\[S_{\infty}=\sum_{i=1}^{\infty}20(-\frac{1}{2})^{n-1}= \lim_{n\rightarrow \infty}\frac{40}{3}(1- (-\frac{1}{2})^n)=\frac{40}{3}.\]

(iii)
\[|S_{\infty}-S_n|=|\frac{40}{3}-\frac{40}{3}(1- (-\frac{1}{2})^n)|=|\frac{40}{3}(-\frac{1}{2})^n)|<0.001\]
\[\Rightarrow  |(\frac{1}{2})^n)|<0.001*\frac{3}{40}=\frac{3}{40000}\]
\[\Rightarrow \frac{1}{2^n}<\frac{3}{40000}     \Rightarrow    2^n>\frac{40000}{3}\]
\[\ln (2^n) >\ln(\frac{40000}{3})   \Rightarrow  n\ln(2)>\ln(\frac{40000}{3})\]
\[\Rightarrow n > \frac{\ln(\frac{40000}{3})}{\ln (2)}\approx 13.70.\]
Hence the least value of $n$ is $14$.