Yes, there are such series. Consider, for example, a sequence such as
$$1, 2, \frac 1 2, 1, \frac 1 4, \frac 1 2, \frac 1 8, \frac 1 4, \dots$$
The series
$$1 - 2 + \frac 1 2 - 1 + \frac 1 4 - \frac 1 2 + \frac 1 8 - \frac 1 4 + \dots$$
is alternating and (absolutely) convergent, but it clearly fails to be monotonically decreasing.
The point, which isn't made often enough (in my opinion) in classes on the subject, is that the convergence of a series is a limit process. In this case, what that means is that the question of convergence is completely determined by the behavior of the series for say $n > N$ for any fixed, finite $N$. If I take a convergent series $\sum a_n$, and I cut off the first 55 quintillion terms, and replace them all with $n!$ to get a new sequence
$$ b_n = \begin{cases} a_n \text{ if } n > 55,000,000,000,000,000,000 \\ n! \text{ if } n \leq 55,000,000,000,000,000,000 \end{cases}$$
Then the sum $\sum b_n$ is still convergent. In fact, $b_n$ is convergent if and only if $a_n$ is. Of course, the sums will be different, but by precisely
$$ \sum\limits_{n=1}^{\infty} b_n - \sum\limits_{n=1}^{\infty}a_n = \sum\limits_{n=1}^{55,000,000,000,000,000,000} (n! - a_n)$$
Which is just a finite number. Of course, I've contrived the example to be huge and ridiculous. But the the point to be made about the integral test is that as long as the function behaves in the desired way beyond some large $N$ (say 55 quintillion), the argument still works for the infinite part of the sum, beyond that, and that is where all problems of convergence lie. Everything else is just addition.
Best Answer
No, there isn't. As long as you can apply the integral test, then it has to work: if $f : [0, \infty) \to [0, \infty)$ is decreasing, then the series $\sum f(n)$ converges iff the improper integral $\int_0^\infty f(t) dt$ converges. This isn't, say, like the root test or the ratio test that can be inconclusive, even when you can apply it (when $\lim a_{n+1}/a_n = 1$ or $\limsup |a_n|^{1/n} = 1$).
But there are two caveats here. First, this only works for nonnegative, decreasing functions $f$. If your function takes negative values, or if it is not decreasing, then you cannot apply the integral test. In that sense, it is not a "silver bullet". Another caveat is that you need to compute an integral to apply the test (or at least determine if it's convergent), and this is not always easy or even feasible. Determining whether $\int_0^\infty f(t) dt$ converges can be just as hard as determining if $\sum f(n)$ converges.
But if you satisfy the hypotheses and if you can determine an explicit antiderivative or otherwise determine the convergence of the integral, then the integral test is a powerful tool.