it syas approxiamate the sum of the series to four decimal places.. do i just keep on testing out numbers and then taking the average or what?
EDIT
Given that $\displaystyle a_n\ge a_{n+1}>0~\&~(a_n)\to 0$ then $\displaystyle \sum\limits_{n = 1}^\infty {( - 1)^n a_n } $ converges.
If $\displaystyle S=\sum\limits_{n = 1}^\infty {( - 1)^n a_n } $ then $\displaystyle \left| {S - \sum\limits_{n = 1}^N {( - 1)^n a_n } } \right| \leqslant a_{N + 1} $.
Since this is an alternating series, the sum to "N+2" will always lie between the sum to "N" and "N+1". That means that the full sum will lie between the sum to N and N+1. And that means that the error will be no larger than the absolute value of the difference between those two sums which is just the absolute value of the N+ 1 term:
$\displaystyle \frac{N+1}{7^{N+1}}$. (I see now that is what Plato just said.)
In order to get the sum correct within 4 decimal places, that is, with error between -0.00005 and 0.00005, you must use N such that $\displaystyle \frac{N+1}{7^{N+1}}< 0.00005$