The asymptotic expansion, in powers of 1/z, of the exponential integal function (Subscript[E, 1])(x) to n terms:

The expansion to n terms is

expint[n_, z_] := (Exp[-z]/z) Sum[(-1)^k  k! / z^k, {k, 0, n}]

The difference compared with the actual result (i.e. the error) for z = 8, for n between 0 and 15, is

diff = N[Table[ expint[n, z] -  ExpIntegralE[1, z], {n, 1, 15}] ]/. z ->8

For technical reasons connected with ListPlot on this version of Mathematica (6.0.1) I multiply by (10^6):

diff2 = diff * (10^6);

Now we plot the error as a function of n (it is useful to plot both the points and the lines between the points; this can be done fairly easily in version 6 as follows:)

ListPlot[{diff2, diff2}, PlotRange ->{{0, 16}, {-1.3 , 0.6}}, Joined -> {True, False}, AxesLabel -> {"n", "error (x (10^6))"}]

Remember this is for z = 8,  We see that the error initially decreases as n increases, but then increases. As n -> ∞, the series diverges.

The actual value of (Subscript[E, 1])[8] is equal to

N[ExpIntegralE[1, 8]]

i.e. about 3.7 x (10^(-5)). Hence the minimum error, 4.8 x (10^(-8)), which is for n = 7, is about a thousand times smaller, so stopping at at this value of n gives a good approximation to the function.

Empirically, the error is least if we stop at the value of n for which the term in the series is least in magnitude.

For asymptotic series: 

(i) if we fix the number of terms in the series, to n say, and let the expansion parameter tend to zero, we get a very good approximation to the function, in which the error is (often) given, to a good approximation, by the first omitted term in the series.

(ii) However, if we fix the value of the expansion parameter and let the number of terms tend to infinity, the series diverges.

For a convergent series (i) is true, but (ii) is not (the series converges as the number of terms tends to infinite, assuming that the expansion parameter is within its radius of convergence.