The integral decides, and it is not the sum BC only

Let $f$ be a function with $f(n) = a_n$. The integral test applies when, on some interval $[N, \infty)$, $f$ is

1. positive,
2. continuous, and
3. decreasing.

Then $\displaystyle\sum_{n=N}^{\infty} a_n$ and $\displaystyle\int_{N}^{\infty} f(x)\,dx$ both converge or both diverge.

Each hypothesis earns its place. Positive and decreasing are what make the rectangle comparison in the next section valid; drop either and the areas no longer trap each other. Continuous is what makes the integral exist. Applying the test to $\sum \frac{\cos n}{n}$ fails at the first hurdle, since those terms change sign, and applying it to $\sum \frac{\ln n}{n}$ requires noticing that $\frac{\ln x}{x}$ increases until $x = e$ and only decreases after.

That last case shows why the word is eventually. A finite number of terms cannot change whether a series converges, so the hypotheses only have to hold from some point onward. Start the integral where the behaviour starts.

Draw $f$ decreasing, and put a rectangle of width $1$ and height $f(n)$ on each interval $[n, n+1]$. Because $f$ is decreasing, $f(n)$ is the largest value of $f$ on that interval, so each rectangle contains the region under the curve there:

$$\sum_{n=1}^{k} f(n) \;\ge\; \int_{1}^{k+1} f(x)\,dx.$$

Slide the rectangles one step left, giving each height $f(n+1)$, and the inequality reverses because now each rectangle sits under the curve:

$$\sum_{n=2}^{k+1} f(n) \;\le\; \int_{1}^{k+1} f(x)\,dx.$$

Together these trap the partial sums between two multiples of the same integral. If the integral is finite the partial sums are bounded above and increasing, so they converge; if it is infinite they are pushed to infinity too. The test is a comparison, and the integral is the thing being compared to.

This is the error worth the most care, because the work is done correctly and only the last line is wrong. Take $\sum_{n=1}^{\infty} \frac{1}{n^{2}}$:

$$\int_{1}^{\infty} \frac{dx}{x^{2}} = \lim_{b \to \infty}\left[-\frac{1}{x}\right]_{1}^{b} = 1.$$

The integral converges, so the series converges. The sum of the series is not $1$. It is $\frac{\pi^{2}}{6} \approx 1.6449$, a fact well outside this course to prove and entirely inside it to use as a warning.

The two inequalities above say exactly how the numbers relate. With $a_1 = 1$:

$$\int_{1}^{\infty} f \;\le\; \sum_{n=1}^{\infty} a_n \;\le\; a_1 + \int_{1}^{\infty} f, \qquad\text{that is}\qquad 1 \le 1.6449 \le 2.$$

So the integral brackets the sum without ever equalling it. The rectangles and the curve have the same behaviour and different areas, and the gap between them is precisely the difference the test throws away in exchange for being easy to compute.

A complete answer has four parts, and omitting the first is what CA4 records:

1. State the function and verify the three hypotheses on the interval you are using. One line each is enough, but they must be there.
2. Set up the improper integral with a limit, not with $\infty$ as a limit of integration.
3. Evaluate it and say whether it converges.
4. Conclude about the series, and say nothing about its value.

The test is at its best on terms containing $\ln x$, or any $f$ you can actually antidifferentiate: $\sum \frac{1}{n \ln n}$ is settled in two lines by $u = \ln x$, and by nothing else in this unit. It is at its worst on factorials, which have no antiderivative to reach for, and on anything with $(-1)^{n}$, which fails positivity outright.