The Integral Test states
Assume $f$ is continuous, positive, and decreasing on [$1, \infty$).
If $\int_1 ^{\infty}f(x)\,dx$ exists and is finite, then $\sum f(n)$ converges and vice versa.
Wouldn't the theorem still hold if $f$ were negative, continuous and increasing?
My attempt to prove this case:
Choose partition P of integers, and $k \; \epsilon \;(x_{i-1},x_i)$ .
Then, $\int _1^\infty f $ is an upper bound for $\Sigma_{k = 1}^{\infty} f(k) $ and a lower bound for $\Sigma_{k = 2}^{\infty} f(k) $.
So $\int_1 ^{\infty}f(x)\,dx$ and $\Sigma_{k = 1}^{\infty} f(k) $ are bounded together, and hence converge/diverge together.
Sure, it would also work if $f$ is negative, continuous, and increasing, but that's so similar to the case of positive, continuous, and decreasing that it really doesn't need to be mentioned (just replace $f$ with $-f$).
The more elegant formulation of the theorem would simply demand that the function is monotone and continuous. But the end usefulness of all three theorems would be more or less the same.