When a series test gives no information
A student must determine whether Σ (from n = 1 to ∞) of a_n converges, where a_n = (n^2 + 1)/(n^3 + n). They apply the ratio test: a_{n+1}/a_n → 1, so they conclude the ratio test is inconclusive, then apply the comparison test against 1/n, noting a_n ≈ n^2/n^3 = 1/n, and conclude the series diverges. A classmate objects that a_n is not actually greater than 1/n for all n (at n = 1, a_1 = 1 while 1/1 = 1; at n = 2, a_2 = 5/10 = 0.5 while 1/2 = 0.5; at n = 3, a_3 = 10/30 ≈ 0.333). The classmate says the comparison is therefore invalid and the series might converge.
1. Confirm the ratio test conclusion and explain why the comparison with 1/n needs care.
Read the worked answer
Ratio test: a_{n+1}/a_n = [(n+1)^2 + 1](n^3 + n) / [(n+1)^3 + (n+1)](n^2 + 1). The numerator and denominator are both degree 5 polynomials in n with leading coefficient 1, so the ratio → 1. The ratio test is indeed inconclusive - it never decides when the limit is exactly 1.
The direct comparison a_n ≥ 1/n is not automatic. Computing: a_n - 1/n = [n(n^2 + 1) - (n^3 + n)]/[n(n^3 + n)] = [n^3 + n - n^3 - n]/[...] = 0. So in fact a_n = 1/n exactly for every n. The classmate's arithmetic (1, 0.5, 0.333) is consistent with equality, not with a_n < 1/n. The comparison is valid - but only because the terms are literally equal, which the student should have noticed.
2. Give a complete, rigorous argument that the series diverges, using a limit comparison test.
Read the worked answer
Take b_n = 1/n. Then
a_n/b_n = [(n^2 + 1)/(n^3 + n)] · n = (n^3 + n)/(n^3 + n) = 1.
Since lim (a_n/b_n) = 1, a finite positive limit, the limit comparison test says Σa_n and Σb_n share the same convergence behaviour. The harmonic series Σ1/n diverges, so Σa_n diverges.
This is cleaner than direct comparison because it sidesteps the need to prove a term-by-term inequality. The limit comparison test is the right tool whenever a_n behaves like a known p-series term.
3. The classmate proposes instead comparing a_n with 1/n^2, arguing that since the numerator has n^2, the series should behave like Σ1/n^2 and converge. Diagnose this reasoning and state the correct heuristic.
Read the worked answer
The heuristic is wrong: you compare the whole term to a p-series, not the numerator to a p-series. Here a_n = (n^2 + 1)/(n^3 + n) ~ n^2/n^3 = 1/n, which is p = 1.
The correct rule: for a rational expression in n, the dominant power in the numerator minus the dominant power in the denominator gives the effective exponent. Here 2 - 3 = -1, so the term behaves like 1/n^1, and since p = 1 ≤ 1 the series diverges.
Formally, a_n/(1/n^2) = n(n^2 + 1)/(n^3 + n) = (n^3 + n)/(n^3 + n) = 1 again - the limit is finite and positive, so limit comparison would say Σa_n behaves like Σ1/n^2, which converges. That is a contradiction, so something must be wrong. The error: a_n/(1/n^2) = a_n · n^2 = n^2(n^2+1)/(n^3+n) = (n^4 + n^2)/(n^3 + n) → ∞, not 1. The classmate divided by the wrong power. Since the ratio blows up, limit comparison with 1/n^2 gives no conclusion; the correct comparison is with 1/n.