A student encounters the series Σ (n! / nⁿ). Which test should they reach for first after checking the divergence test, and why?
AThe integral test, because it works for positive decreasing functions
BThe ratio test, because n! in the numerator makes the ratio aₙ₊₁/aₙ simplify cleanly
CThe comparison test, since n! grows faster than nⁿ for large n
DThe root test, because the nth root of n! is easy to evaluate
The ratio test is designed for series where factorials or exponentials appear, because taking aₙ₊₁/aₙ causes those terms to cancel nicely. For n!/nⁿ, the ratio becomes [(n+1)!/(n+1)^(n+1)] / [n!/nⁿ] = (n/(n+1))ⁿ → 1/e < 1, confirming convergence. The integral test (option A) would be extremely difficult to apply here. Reading the algebraic structure — factorial present — immediately points to the ratio test.
Question 2 Multiple Choice
A student uses the alternating series test to prove that Σ ((-1)ⁿ/√n) converges, then concludes the convergence is absolute. What error has been made?
ANo error — if the alternating series test proves convergence, convergence is absolute
BThe alternating series test cannot be applied here because √n is not an integer
CThe alternating series test only establishes conditional convergence; absolute convergence requires testing Σ |aₙ| = Σ 1/√n separately, which is a divergent p-series (p = 1/2)
DThe student should have used the ratio test instead, which directly determines absolute convergence
The alternating series test only guarantees conditional convergence — the original series with alternating signs converges, but nothing is said about convergence of the absolute values. Σ 1/√n is a p-series with p = 1/2 < 1, which diverges. So Σ ((-1)ⁿ/√n) is conditionally convergent but NOT absolutely convergent. This matters: conditionally convergent series can be rearranged to sum to any value (Riemann rearrangement theorem), while absolutely convergent ones cannot.
Question 3 True / False
The divergence test should always be the first test applied to any series, regardless of structure.
TTrue
FFalse
Answer: True
The divergence test costs almost nothing — just evaluate limₙ→∞ aₙ — and immediately resolves any series whose terms don't approach zero. If the limit is nonzero, the series diverges and no further analysis is needed. If the limit is zero, the test is inconclusive and you move on. Given this asymmetry (instant conclusion vs. no cost), always applying it first is the correct strategic discipline.
Question 4 True / False
If the ratio test yields L = 1, the series definitely converges.
TTrue
FFalse
Answer: False
When the ratio test yields L = 1, the test is inconclusive — it gives no information about convergence or divergence. Both the harmonic series Σ 1/n (diverges) and Σ 1/n² (converges) yield L = 1 under the ratio test. In this case you must switch tests — typically limit comparison with a p-series for rational-function terms.
Question 5 Short Answer
Why does recognizing a series' algebraic structure matter for choosing a convergence test — why not just try tests in a fixed order every time?
Think about your answer, then reveal below.
Model answer: Different tests exploit specific algebraic structures: factorials and exponentials make the ratio test's cancellation work; nth powers collapse under the root test; rational functions compare naturally to p-series. Applying tests randomly leads to intractable calculations — trying the integral test on a factorial series is essentially impossible. Reading the structure of aₙ first directs you to the test where the algebra simplifies.
Strategic test selection is not just faster — it is often the difference between a solvable and an unsolvable computation. The tests were designed to exploit specific features, so matching the test to the algebraic form is a mathematical insight, not just a time-saving habit.