Verified answer

This is a great question. It might be instructive to formulate a statement that sort of "ought to be true" under your intuition, and then see if it is true. As an example, let's try the following:

"If f is continuous on [1, oo), and the integral from 1 to infinity of f(x) dx converges, and a sequence (a_n) satisfies a_n = f(n) for all positive integers n, then sum a_n converges."

Is that true? Well, no, it isn't. Consider a function f(x) defined by constructing its graph as follows. For each positive integer n:

- Draw a straight line segment in the plane from the point (n - 1/2^n, 0) to (n,1).

- Draw another straight line segment in the plane from the point (n,1) to (n + 1/2^n, 0).

- Draw a straight line segment from (n + 1/2^n, 0) to (n+1 - 1/2^(n+1), 0).

The result of this process is the graph of a continuous function defined on [1/2, infinity). The graph has a little "spike" at each integer value n = 1, 2, 3, ... and the spikes get narrower as n gets larger. Aside from these spikes, f(x) is 0. [You could give a piecewise-defined algebraic rule for f(x) in terms of x, but having those formulas doesn't really help, unless you like looking at formulas.]

The nth "spike" is made up of two triangles with base 1/2^n and height 1; so the area under the nth "spike" is precisely 1/2^n. Using the fact that the geometric series sum 1/2^n converges, it is relatively straightforward to show from the definitions that the integral from 1 to infinity of f(x) dx is convergent. You can even compute the value of this integral. It is 3/4. [Not the full infinite series 1/2 + 1/4 + 1/8 + ... = 1, because if we integrate f(x) from 1 to infinity, you miss half of the first spike. So the full geometric series is 1/4 more than the integral.]

But by construction f(n) = 1 for all n. And the series sum_{n=1 to infinity} 1 clearly diverges. So the general statement written above cannot be true.

Note that in this example the sequence a_n in question (namely, a_n = 1) *was* nonincreasing--- it's the function f(x) that wasn't.

This is really the key to the whole thing. If a function f(x) is nonincreasing, then [by arguments you are familiar with] there are a simple and easy to write down relationships between integrals of f(x) on intervals of the form [1, b) and numerical sums of the form f(1) + f(2) + ... + f(k). And these relationships are what allow you to go back and forth between the convergence of the integral of f(x) on [1, infinity) and the convergence of the numerical series f(1) + f(2) + ... + f(n) + ...

If f(x) is not assumed to be monotonic, "spikelike" examples of the sort given above show you that there is no reason to expect *any relationship at all* between integrals of f(x) on intervals, and sums of the form f(1) + f(2) + ... + f(n), and hence no reason to expect anything like the integral test. The fundamental reason is that integrals of f(x) have to do with the behavior of f(x) on actual intervals, and the sum f(1) + f(2) + ... depends only on the values of f on a discrete set of points. Without monotonicity, there doesn't have to be any relationship between values of f(x) at the endpoints of an interval, and the values of f(x) for x inside that interval, so there is no relationship between the integral and the series. As examples with narrow spikes show, with non-monotonic f, you can change what the numbers f(n) are, n = 1, 2, 3, ..., almost arbitrarily, without affecting the integral of f(x) all that much. [If you alter the idea of the above construction just a little, then for _any_ sequence of real numbers (c_n) whatsoever, you can construct a continuous function g(x) on [1, infinity) with the property that g(n) = c_n holds for all n, and also with the property that the integral from 1 to infinity of g(x) dx converges.]

Now, the integral test as usually stated, is by no means "the most general possible test involving integrals and sums." You could certainly weaken monotonicity to something slightly less than it (so that the sequences f(n) that the test applied to, were no longer necessarily monotonic). But whatever assumption you made, it would have to be complicated enough to get around the general phenomenon above. Without monotonicity (or something close to it), there is no relationship at all between the integrals of f(x) and the sums of values of f(1) + f(2) + ... f(n). Even if all you care about is convergence or divergence, and not the actual values (if any).

I hope this helped.