Had this epiphany while I was thinking about the end behavior of the oh-so-special logarithm. The natural logarithm diverges, but at a very slow rate; proof of this can be found in its derivative, 1/*x*, whose discrete series, the harmonic series, diverges.

Well, so all of this lends to thinking about the integral test. It states the following (copied from Wikipedia):

Consider an integer

Nand a non-negative monotone decreasing functionfdefined on the unbounded interval [N, ∞). Then the series

converges if and only if the integral

is finite. In particular, if the integral diverges, then the series diverges as well.

Damnit, I wish I had math typing software here. Anyway, let’s start with some definitions:

Let f(x) be an arbitrary function defined, continuous, nonnegative, and decreasing for all x > 0, and let F(x) be defined and continuous on the same interval. If f(x) were increasing the divergence would be clear; if the range was extended to all real numbers then…. iono… the universe would implode.

Let dy/dx = y’ ≡ f(x), and the (antiderivative/indefinite integral of y’) = y ≡ F(x);

Let S(f, a, b, x) ≡ the definite integral of f with respect to x, evaluated from a to b. [since S looks like an integral sign]

Let lim(f, x, c) ≡ the limit of f as x approaches c.

From the fundamental theorem of calculus we know the following:

S(f(x), a, b, x) = F(b) – F(a).

And if the upper limit is to be infinite, that is, an improper integral is involved:

S(f(x), a, ∞, x) = lim(F(b), b, ∞) – F(a).

Intuitively, if the above improper integral were to diverge, then:

lim(F(b), b, ∞) = ∞.

[F(a) could not be infinite due to the restrictions in the definitions; thus F(b) has to be infinite.] Again, intuitively, if this is the case then

lim(F'(b), b, ∞) = lim(f(b), b, ∞) ≠ 0.

Because to be diverging, the function has to be “moving” (i.e. slope not nearing 0) as b -> ∞ (if the slope was closing in on 0, the function would be staying constant, converging to a real number).

From the divergence test we know that a discrete sequence f*(n) corresponding to its infinite series and to f(x) must diverge if and only if lim(f*(n), n, ∞) ≠ 0. This can be extended to the corresponding function; if lim(f*(n), n, ∞) ≠ 0, then clearly also lim(f(x), x, ∞) ≠ 0 (and vice versa). And this result is of course what we have above from our calculus analysis. So it follows that if the improper integral diverges, so does the infinite series.

This is only a half-proof because the inverse of the divergence test does not follow (e.g. harmonic series, which is what made me start thinking of this in the first place). Maybe I’ll have another epiphany someday.