Originally Posted by

**mark1950** Given the geometric series, $\displaystyle 1,-\frac{1}{2},\frac{1}{4}...$ find the least number of n so that the difference between the Sum to infinity and the Sum to n is less than $\displaystyle 10^{-5}$

This is my take:

$\displaystyle r = -\frac{1}{2}$

$\displaystyle a = 2$ Should be a = 1, but that doesn't really affect the calculation much.

Substituting r and a into the formula of Sum to infinity and Sum to n,

$\displaystyle \left|\frac{2}{1+\frac{1}{2}} - \frac{2[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}\right|< 10^{-5}$

$\displaystyle \left|\frac{4}{3} - \frac{2[1-(-\frac{1}{2})^n]}{\frac{3}{2}}\right| < 10^{-5}$

$\displaystyle \left|\frac{4}{3} - \frac{4}{3}\Bigl[1-\Bigl(-\frac{1}{2}\Bigr)^n\Bigr]\right| < 10^{-5}$

$\displaystyle \left|\frac{4}{3}\Bigl[1-\Bigl[1-\Bigl(-\frac{1}{2}\Bigr)^n\Bigr]\Bigr]\right| < 10^{-5}$