The iteration should be simple; it's how you know when you're done, without knowing the answer, that's trickier. Basically, if you need to know the answer to 5 decimal places, then stop as soon as the 5th decimal place hasn't changed in some arbitrary number of trials.
A more elegant way (not requiring you to multiply by 4 after every step to ensure the precision of the final number is within the epsilon) is to quit as soon as you reach a value that does not change the least significant digit of the decimal value that would produce the correct value of Pi. Pi to 5 decimals is 3.14159. Divided by 4 (producing the necessary sum of the infinite series), that's .7853975. This is what you have to get to. Therefore, the largest fraction that can be summed to the series that will not affect the precision is 1E-8 (one hundred millionth).
So, the ending point is the iteration N at which $|\dfrac{-1}{2n-1} + \dfrac{1}{2n+1}| < .00000001$, so that at the end of the iteration (one iteration being a subtraction and an addition), the series sum has not changed by a greater degree than the epsilon. By inspection, we see that the sum of the two fractions will be negative, so:
$\dfrac{1}{2n+1} - \dfrac{1}{2n-1} = -0.00000001$
$\dfrac{2n-1}{(2n+1)(2n-1)} - \dfrac{2n+1}{(2n+1)(2n-1)} = -1*10^{-8}$
$\dfrac{(2n-1)-(2n+1)}{(2n+1)(2n-1)} = -1*10^{-8}$
$\dfrac{-2}{4n^2-1} = -1*10^{-8}$
$-2 = (-1*10^{-8})(4n^2-1)$
$\dfrac{-2}{-1*10^{-8}} = 4n^2-1$ <-- from here you can pretty much plug in any needed precision
$2*10^{8} = 4n^2-1$
$200000001/4 = n^2$
$50000000.25 = n^2$
$n > 7071$
Plugging this back in to check, we get $\dfrac{1}{2(7072)+1} - \dfrac{1}{2(7072)-1} = \dfrac{1}{14145} - \dfrac{1}{14143} = -0.9997*10^{-8}$ which is less significant than $-1*10^{-8}$.
So, it should take 7072 iterations, at least, of subtracting the inverse of an odd natural number, then adding the next smaller odd inverse, to arrive at a sum that will produce pi to the necessary precision. In fact, it may actually take many more to produce a value that rounds correctly to the necessary number of decimal places; if you simply truncate to the needed sig figs, this is the answer.