You're essentially missing the relativistic velocity addition law. Velocities don't add linearly, rapidities do (in co-linear problems), because it is the rapidity parameter that multiplies a boost generator to get the logarithm of the relavant Lorentz transformation, not velocity. More intuitively: they don't add linearly, because you're adding together the same velocity increments (as you observe), but in different frames of reference, so that time dilations and all the rest of it have to be applied to the same interval as you go.
In a one dimensional problem, the relevant Lorentz transformation (see notes at end) is:
$$\Lambda(\eta) = \exp\left(\eta\,\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\right)\tag{1}$$
where $\eta$ is the boost's rapidity. Suppose you repeatedly impart the same transformation to yourself, which is what you're describing: there is the same impulse each time, and, as you say, the relevant inertia for your problem is your rest mass. Then after $n$ such identical moves, your overall transformation is $\Lambda(\eta)^n=\Lambda(n\,\eta)$. So, each boost adds a rapidity:
$$\eta = \operatorname{artanh}\left(\frac{v}{c}\right)\tag{2}$$
and your final velocity is thus:
$$v_n = c\,\tanh(n\,\eta) = c\,\tanh\left(n\,\operatorname{artanh}\left(\frac{v}{c}\right)\right)\tag{3}$$
So, in your example, if $v=0.2\,c$ is the relative velocity imparted by each boost, your velocity after five of them is $c\,\tanh\left(5\,\operatorname{artanh}\left(\frac{1}{5}\right)\right)\approx0.767\,c$
(1) makes obvious the claim that the combination of any finite number of finite boosts, no matter how big, is always equivalent to a boost of finite rapidity, thus (through (2)) less-than-$c$ velocity and thus the accelerated object will still have finite total and kinetic energy.
Exercise:
Work formulas (2) and (3) out from (1). The following steps are hints:
- Expand the matrix exponential for the Lorentz transformation in (1);
- Using this expression, read off the distance travelled by the origin of one co-ordinate system ($A$) relative to the other ($B$) over a unit time as measured by $B$ to find expressions for the Lorentz transformation in terms of relative velocity;
- Derive (2) and (3) by comparing the two expressions you now have.
Actually something like Mathematica will crunch it for you: remember to use MatrixExp, not Exp, when you're doing a matrix exponential (Exp, like any like function such as Sin, Cos, ... simply calculate elementwise exponentials / relevant function for matrices. Something similar holds for Sage and Maple, too).
The form of the Lorentz transformation in (1) is justified by basic assumptions about the universe's symmetries and homogeneities (see references at the end). Continuity assumptions then show that the Lorentz transformations must form a matrix Lie group, so, in one spatial dimension, we must have a one parameter group and therefore there must be some rapidity parameter that adds linearly with combined transformations: one then simply has to discover this parameter's relationship with the distance over time velocity. The matrix in the one parameter Lie algebra (the matrix inside the exponential) is constrained to the form shown because it can't be a rotation (rotations can switch the sign of the time co-ordinate and thus violate causality) and it can't have on-diagonal elements just because they give unphysical results.
See:
Palash B. Pal, "Nothing but Relativity," Eur.J.Phys.24:315-319,2003
Jean-Marc Levy-Leblond, "One more derivation of the Lorentz transformation"
