What kind of series is the following what kind of series is the following and can I get a generalized formula for it?
$T_1 = (1-\frac{1}{2N^3})$ 
$T_2 = (1- T_1.\frac{1}{2N^3}) $ 
$T_3 = (1 - T_2.\frac{1}{2N^3})$ 
$T_4 = (1 - T_3.\frac{1}{2N^3})$ 
..
.
$T_x = (1 - T_{x-1}.\frac{1}{2N^3})$ 
 $ P_{total} =  T_1 + T_2 + T_3 + ... T_x $

You have a recurrent formulation of a series. Let $a=\frac1{2N^3}$. You can see that the explicit formula
$T_n = \frac{1 + (-a)^n a}{1+a}$ solves your given recurrence $T_{n+1} = 1 - a T_{n}$, with $T_0=1$, because
$$ 1 - a T_n = 1 - \frac{1 + a - a - a\cdot(-a)^n\cdot a}{1+a} = \frac{1 + (-a)^{n+1} a}{1+a} = T_{n+1}$$
and of course $T_0 = \frac{1 + (-a)^0 a}{1+a} = 1$.

For your $P_\mathrm{total}$ you have
$$P_\mathrm{total} = \sum_{n=1}^x T_n = \frac{\sum_{n=1}^x1 + a\sum_{n=1}^x(-a)^n}{1+a} = \frac{x + a\left(\frac{1-(-a)^{x+1}}{1-(-a)} - 1\right)}{1+a} = \frac{x + a \biggl(\bigl((-a)^x - 1\bigr) a + x\biggr)}{(1 + a)^2}.$$
by using the formula for a geometric sum.
