Part CCSIR NET December 2025a-terminating-decimal-is-approached-from-below-by-borrowing-one-digit-and-trailing-nines

A terminating decimal is approached from below by borrowing one digit and trailing nines

Let $\alpha = \sum_{l=1}^{2025}$ $a_{l}/10$ ˡ with $a_{l} \in$ {0,1,…,9} for all 1≤l≤2025 and $a_{2025}\ne0$ . Let $(\beta_{n})_{n}\ge1$ be a strictly increasing sequence of positive real numbers in (0,1) that converges to $\alpha$ . For each n≥1, write $\beta_{n} = \sum_{l=1}^\infty$ $b_{n,l}/10$ ˡ with $b_{n,l}$ ∈{0,1,…,9}, for the infinite decimal expansion of $\beta_{n}$ . Which of the following statements are necessarily true?

A.There exists a positive integer N such that for all n≥N, $b_{n,l}$ $= a_{l}$ for all 1≤l≤2023.✓
B.There exists a positive integer N such that for all n≥N, $b_{n,2025}$ $= a_{2025} - 1$ .✓
C. $\beta_{n}$ is rational for infinitely many n.
D.There exists a positive integer N such that for all n≥N, $b_{n,2024}$ $= a_{2024} - 1$ .

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: aₙ^{1/n} → L ⇒ aₙ₊₁/aₙ → L

A terminating decimal is approached from below by borrowing one digit and trailing nines

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