Are the higher-order terms in the series for energy really negligible? To show that energy in special relativity reduces to $E=m+mv^2/2$ for low velocities, if we make a Taylor expansion of $m\gamma$ around $v=0$ we get $$E=m+mv^2/2+3mv^4/8+\cdots$$ But why can we cutoff after the second term? How does one know all the other terms don't add up to something significant?
We get $E = \text{rest energy}+\text{kinetic energy}+\frac{3mv^4}{8} + \cdots$ What kind of energy is the $3mv^4/8$ term?
 A: Because the radius of convergence of the expansion is 1, so you know the Taylor series is good until a sizable fraction of v=1 (units where c=1). The heuristic for Taylor series is that for an analytic function (or one with poles and cuts), your expansion is good in a circle in the complex plane which extends until the first singularity. Further, inside the circle, away from the edge, the convergence of the Taylor series is like the convergence of a geometric series, the error is about the same size as the first neglected term. The more precise statement is that the size of the consecutive terms are approximately reduced by the ratio of the distance from the center to the distance to the first singularity, this is the comparable geometric series.
So in this case, for $1\over \sqrt{1-v^2}$, you have a branch cut starting at $v=1$, so starting at v=0, the radius of convergence is 1, and the error for small v in taking the first two terms is just well estimated by the third term. You are far from the cut.
But in this case, you can also show it explicitly without complex analysis ideas, just by estimating the difference between the function and the first two terms in the expansion using Taylor's formula (derived by repeated integration by parts):
$$ |{1\over \sqrt{1-v^2}} - 1 - {v^2\over 2} | = | \int_0^v {d^3\over du^3} ({1\over\sqrt{1-u^2}}) {u^2\over 2} du |$$
and then note that the third derivative of the function in question is bounded in the given interval by, say, 10, so that the error is at most 10 times the neglected term.
A: 
We get E=rest energy+kinetic energy+3mv48+⋯ What kind of energy is the
  3mv4/8 term?

Not quite.  That should be:
E = invariant energy + (relativistic) kinetic energy
In other words, what you've labeled "kinetic energy" is actually just "Newtonian kinetic energy".
Once you see this, your question evaporates.
