Turn a positive tail into an integral comparison

Try

Compare the terms n/(n+1) and 1/n². The first approach 1, so continuing to add them cannot produce a finite total. The second approach zero, but that fact alone does not settle their sum. Predict which test provides an immediate answer and which needs additional tail information.

Predict

Sketch a positive decreasing curve f(x) with sample heights f(n). Rectangles based on integer intervals lie on opposite sides of the curve depending on whether the left or right endpoint supplies the height. Predict which improper integrals will bound the tail after the Nth term.

Experiment

For f(x)=1/x², compare the series tail R_N=Σₙ₌N+1∞1/n² with the neighboring integral tails.

N	Lower bound 1/(N+1)	Upper bound 1/N
5	1/6	1/5
10	1/11	1/10
20	1/21	1/20

Draw one right rectangle over [N,N+1] and one left rectangle over [N+1,N+2]. Use those pictures to explain the shifted lower endpoint. These bounds concern the omitted tail, not the full sum, and neither is generally its exact value.

Understand

The nth-term divergence test says that if aₙ does not approach zero, Σaₙ diverges. Failure to meet this necessary condition can be caused by a nonzero finite limit, unbounded terms or persistent oscillation. If aₙ does approach zero, the test is inconclusive; a different test must address how rapidly or with what cancellation the tail accumulates.

The standard integral test applies when aₙ=f(n) and f is positive, continuous and decreasing from some index onward. Under these hypotheses, the series and the improper integral of f have the same convergence behavior. The proof compares positive rectangular areas with the area under the decreasing curve. A finite number of early exceptions does not affect convergence, but must be retained when calculating an actual total.

For x^(−p) with p>0, positivity, continuity and decrease hold on [1,∞). Its tail integral converges exactly for p>1. Therefore the p-series Σ1/nᵖ converges exactly for p>1. For p≤0, the term test gives divergence without using the decreasing-function integral test. At the boundary p=1, the logarithmic integral diverges, matching harmonic-series divergence.

When the integral test establishes convergence, it also gives ∫_(N+1)^∞f≤R_N≤∫_N^∞f. For 1/n², integrating produces 1/(N+1) and 1/N. Adding these bounds to a computed S_N gives an interval containing the full sum. This is more informative than simply noticing that successive partial sums change less.

The series does not equal the comparison integral. For 1/n² the integral from 1 to infinity is 1, but the first two terms of the series already sum to 5/4. The theorem compares convergence and tail sizes; it does not identify rectangles with their underlying curved regions. Similarly, a numerically plausible smooth curve through samples does not supply the required monotonicity or continuity unless its formula or assumptions justify them.

Apply

Check the term limit first, then name every integral-test hypothesis. For remainder bounds, label which terms are omitted and use the correct N or N+1 endpoint. In the p-series proof, handle p=1 and nonpositive p explicitly rather than dividing by a parameter expression that could be zero.

Continue

Next compare unfamiliar terms directly with known p-series or geometric tails. The direction of an inequality determines whether it supplies useful information.