A student evaluates a series and finds that the terms a_n approach 0 as n → ∞. What can they conclude using the Divergence Test?
AThe series converges, since the terms go to zero
BThe series diverges, since vanishing terms cause partial sums to stabilize
CNothing — the Divergence Test is inconclusive when a_n → 0; further analysis is required
DThe series converges absolutely
This is the central trap of the Divergence Test. a_n → 0 is NECESSARY but not sufficient for convergence. The harmonic series Σ 1/n has terms going to zero yet diverges. The Divergence Test can only prove divergence (when a_n does not approach 0); it cannot prove convergence. When a_n → 0, you must use another test.
Question 2 Multiple Choice
What does the Divergence Test conclude about the series Σ n/(2n + 1)?
AThe series converges because n/(2n+1) < 1 for all n
BThe series diverges because lim n/(2n+1) = 1/2 ≠ 0
CThe series diverges because 1/2 < 1
DThe test is inconclusive because the terms are bounded
As n → ∞, n/(2n+1) → 1/2 ≠ 0. Since the terms do not approach zero, the Divergence Test immediately confirms divergence — no further analysis needed. Option A confuses boundedness of terms with convergence of the series. Option D is wrong for the same reason: being bounded does not make the Divergence Test inconclusive; it is inconclusive only when a_n → 0.
Question 3 True / False
The Divergence Test can be used to confirm that a series converges.
TTrue
FFalse
Answer: False
The Divergence Test is strictly one-directional: it can only prove divergence. If a_n does not approach 0, the series diverges. If a_n → 0, the test says nothing — you cannot conclude convergence. To prove convergence, you need a different test (integral test, comparison test, ratio test, etc.).
Question 4 True / False
The harmonic series Σ 1/n diverges even though its terms approach zero.
TTrue
FFalse
Answer: True
This is the definitive counterexample to the misconception that a_n → 0 implies convergence. The harmonic series Σ 1/n diverges (shown by the integral test or Cauchy condensation), yet 1/n → 0. It is famous precisely because it violates naive intuition — terms can vanish without the sum stabilizing.
Question 5 Short Answer
Why is it not enough for a series to have terms approaching zero in order to conclude that the series converges? Give an example that illustrates your answer.
Think about your answer, then reveal below.
Model answer: Terms approaching zero is necessary but not sufficient for convergence. The partial sums must stabilize, which requires the terms to decrease fast enough — and a_n → 0 alone does not guarantee this. The harmonic series Σ 1/n is the classic example: 1/n → 0, yet the series diverges because the terms shrink too slowly. Terms like 1/n² decrease fast enough (series converges); terms like 1/n do not.
The Divergence Test tells you only what is ruled out (divergence is certain when a_n ↛ 0), not what is established. Convergence requires a separate positive argument. Keeping the harmonic series in mind as the standard counterexample helps prevent the 'vanishing terms ⟹ convergence' fallacy.