You apply the ratio test to a series and find L = lim(n→∞) |a_{n+1}/a_n| = 1. What can you conclude?
AThe series converges, since L is not greater than 1
BThe series diverges, since L equals 1 and terms do not shrink
CThe test is inconclusive — the series may converge or diverge
DThe series converges absolutely
L = 1 is the inconclusive case. The ratio test cannot distinguish between convergence and divergence when L = 1. For example, both Σ 1/n (diverges) and Σ 1/n² (converges) give L = 1. This is the most important exception to memorize: L < 1 → absolute convergence; L > 1 → divergence; L = 1 → no information.
Question 2 Multiple Choice
For which series would the ratio test be the most natural and effective choice?
AΣ 1/n² — a p-series where terms decay as a power function
BΣ 2ⁿ/n! — a series involving both an exponential and a factorial
CΣ 1/(n ln n) — a series needing the integral test
DΣ (-1)ⁿ/n — a series best handled by the alternating series test
The ratio test excels when terms involve factorials (n!) or exponentials (aⁿ), because the ratio a_{n+1}/a_n simplifies beautifully. For Σ 2ⁿ/n!, the ratio is 2^(n+1)/(n+1)! ÷ 2ⁿ/n! = 2/(n+1) → 0 < 1, so it converges. P-series and logarithmic series give L = 1 (inconclusive), and the alternating series test is more direct for alternating signs.
Question 3 True / False
The ratio test works by checking whether a series eventually behaves like a convergent or divergent geometric series.
TTrue
FFalse
Answer: True
This is the underlying logic. The ratio test computes L = lim |a_{n+1}/a_n|. If L < 1, terms eventually decrease faster than a geometric series with ratio L (which converges). If L > 1, terms eventually grow like a geometric series with ratio L > 1 (which diverges). The test is literally asking: 'What geometric series does this series asymptotically resemble?'
Question 4 True / False
If the ratio test gives L = 1 for a series, that series is expected to converge slowly but steadily.
TTrue
FFalse
Answer: False
L = 1 is inconclusive — the series may converge or diverge. Σ 1/n gives L = 1 and diverges; Σ 1/n² gives L = 1 and converges; Σ 1/(n ln n) gives L = 1 and diverges. L = 1 provides no information. Interpreting L = 1 as 'borderline convergence' is a fundamental misunderstanding of the test's scope.
Question 5 Short Answer
When applying the ratio test to a series involving n!, what algebraic identity is essential, and why does it make the factorial term tractable?
Think about your answer, then reveal below.
Model answer: (n+1)! = (n+1) · n!. This allows the (n+1)! in the numerator of the ratio to cancel with n! in the denominator, leaving only the factor (n+1). Without this identity, the factorial in the ratio would not simplify cleanly.
For example, in Σ n!/nⁿ, the ratio is [(n+1)!/(n+1)^(n+1)] / [n!/nⁿ]. Applying (n+1)! = (n+1)·n! lets you write this as (n+1)·n! / ((n+1)·(n+1)ⁿ) · nⁿ/n! = (n/(n+1))ⁿ → 1/e < 1. The ability to cancel the factorial is the entire reason the ratio test is the right tool for factorial-containing series.