Compare the tail to a known benchmark

Try

For n≥1, the quantity 1/(n²+1) is smaller than 1/n². Since the larger positive series has a finite sum, predict what must happen to the smaller one. Now compare 1/n² with 1/n. Being smaller than a divergent benchmark is compatible with convergence, so that comparison proves nothing by itself.

Predict

You want to prove divergence of 1/√(n²+1). Decide whether an upper bound or a lower bound by a divergent series would help. A lower bound forces its partial sums to grow at least as much as the benchmark. Write the required direction before manipulating denominators.

Experiment

Use n²+1≤2n² for n≥1 to obtain √(n²+1)≤√2 n. Taking positive reciprocals reverses the inequality, giving 1/√(n²+1)≥1/(√2 n).

Goal	Useful comparison	Benchmark behavior
Prove convergence	0≤aₙ≤bₙ	Σbₙ converges
Prove divergence	aₙ≥bₙ≥0	Σbₙ diverges
Inconclusive	0≤aₙ≤bₙ	Σbₙ diverges

Supply your own pair of series for the third row. Both 1/n and 1/n² fit beneath the same harmonic upper bound but behave differently. This gives a concrete test of why a proposed inference is invalid.

Understand

Direct comparison works with eventually nonnegative terms because their partial sums preserve order. A convergent positive upper series bounds the smaller partial sums, which are increasing and therefore converge. A divergent nonnegative lower series forces the larger sums to be unbounded. Negative or alternating terms require a separate absolute-value or cancellation argument; the positive comparison conclusion cannot simply be transferred without checking signs.

The limit comparison test uses eventually positive aₙ and bₙ with aₙ/bₙ→L for a finite positive L. Choose a point far enough into the tail that the ratio lies between L/2 and 3L/2. Then aₙ is squeezed between positive constant multiples of bₙ, so direct comparison proves that the series converge or diverge together. Finitely many earlier terms affect the total but not tail convergence.

For aₙ=(3n+1)/(n³+2), compare with bₙ=1/n². Their ratio is (3n³+n²)/(n³+2), tending to 3. The benchmark p-series converges, so the given positive series converges. Leading-power intuition suggests the benchmark, but calculating the ratio establishes the needed relationship.

Ratios tending to zero or infinity need a one-sided version of the reasoning rather than the finite-positive equivalence statement. If aₙ/bₙ→0 and Σbₙ converges with positive terms, an eventual upper comparison proves convergence. If that benchmark diverges, the same zero ratio alone does not decide aₙ. Always identify which theorem version is being used and which direction supplies the conclusion.

Comparison also supports bounds. Since 1/(n²+1)≤1/n², its omitted tail after N is at most 1/N by the integral-test bound for the benchmark. That is a guaranteed upper bound, not its actual remainder. Tighter comparisons can improve accuracy without changing the underlying convergence argument.

Apply

In the practice bank, write the goal before choosing a comparison. Keep reciprocals’ sign and inequality reversal explicit. For limit comparison, show eventual positivity and a finite positive limit. Explain why one misleading harmonic upper bound cannot determine either convergence or divergence.

Continue

Positive comparison ignores cancellation. Next analyze alternating sums, where controlled sign changes can produce convergence even when the absolute-value series diverges.