Approximation of $\pi$, with an error of less than $\frac{1}{2}\times 10^{-8} $ This is what I've achieved so far:
$$\tan^{-1}1 = \frac{\pi}{4} \Rightarrow \pi = 4\tan^{-1}1$$
$$\tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5}+ \cdots + (-1)^k\frac{x^{2k+1}}{2k+1}$$
$$\pi = 4\tan^{-1}1 = 4\left(1-\frac{1}{3}+\frac{1}{5} +\cdots + (-1)^k\frac{1}{2k+1}\right)$$
What is the next step?
 A: The remainder for an alternating series is given by $$|R_k| \le b_{k+1}.$$
This means you are looking for a $k$ for which $$\frac{1}{2(k+1)+1} < \frac{1}{2\cdot 10^8}.$$
This tells us that the $k$ we are looking for is greater than $$10^8 - \frac{3}{2}$$ which is fairly large. It would be impractical to actually compute $\pi$ with this series. Heuristically, this is a bad series to use, since it is not absolutely convergent. In this sense it converges "slowly."
There are several better methods out there. There is a book by the Borwein brothers "Pi and the AGM" that discusses approximations of $\pi$. For a long time they held the record for the number of digits of $\pi$ computed. I am not sure if they still hold that record.
A: You have all that you need. You need to solve this for k : $\frac{1}{2}10^{-8}>\frac{4}{2k+1}$
and then comptute the sum to k.
By solving that inequation, you know that every term you add wont get you further from Pi by more than the error you want. 
http://en.wikipedia.org/wiki/Alternating_series
See the part "approximating sums"
But your way of approach is bad because minimal k is very big here.
