The solution to a summation $4^n/(4^n+2 )$ I found this problem on summations, and I'm not really sure how to solve it. Could someone give a hint as to how to do so?
Find the value of
$$\sum_{i=1}^{1000}f\left(\frac{i}{1000}\right),\qquad f(x) = \frac{4^x}{4^x+2}$$ It came on an exam where we couldn't use calculators, and it apparently is an integer answer, though Wolfram Alpha disagrees...(Even if it isn't, I would still like to know how to do it)

$$\sum_{n=1}^{1000}\frac{4^n}{4^n+2} = \sum_{n=1}^{1000}\frac{2^{2n}}{2^{2n}+2}  =  \sum_{n=1}^{1000}\frac{2^{2n-1}}{2^{2n-1}+1}  = \frac23 + \frac89 + \frac{32}{33} + \frac{128}{129} + \cdots + \frac{2^{1999}}{{2^{1999}+1}} \approx 999.5149000482058.$$ 
