Integral Test for Convergence BC only

With $f(n) = a_n$ and $f$ positive, continuous and decreasing on $[N, \infty)$, the series and $\int_{N}^{\infty} f$ converge together or diverge together. All three hypotheses do work: positivity and monotonicity are what make the rectangle comparison valid, continuity is what makes the integral exist, and each only has to hold eventually, since finitely many terms cannot change convergence. That matters for $\sum \frac{\ln n}{n}$, whose function increases until $x = e$, and rules the test out entirely for $\sum \frac{\cos n}{n}$, whose terms change sign. Rectangles of height $f(n)$ on $[n, n+1]$ sit above a decreasing curve and rectangles of height $f(n+1)$ sit below it, which traps the partial sums between two copies of the same integral.

The value of that integral is not the sum of the series, and $\sum \frac{1}{n^{2}}$ settles the point: $\int_{1}^{\infty} x^{-2}\,dx = 1$ exactly, while the sum is $\frac{\pi^{2}}{6} \approx 1.6449$. The two bounding inequalities give $\int_{1}^{\infty} f \le \sum a_n \le a_1 + \int_{1}^{\infty} f$, here $1 \le 1.6449 \le 2$, so the integral brackets the sum and never equals it. A complete answer therefore verifies the hypotheses, sets the improper integral up with a limit rather than an infinite endpoint, evaluates it, and then concludes about convergence only. The test is strongest where an antiderivative exists, which makes $\sum \frac{1}{n \ln n}$ a two-line problem, and useless against factorials or alternating signs.