taiwan-gsat 2022 Q8

taiwan-gsat · Other · ast__math-a 8 marks Sequences and series, recurrence and convergence Convergence proof and limit determination

Suppose two sequences $\langle a_n \rangle$ and $\langle b_n \rangle$ satisfy $b_n + \frac{4n-1}{n} < a_n < 3b_n$ for all positive integers $n$. Given that $\lim_{n \to \infty} a_n = 6$, select the correct options.
(1) $b_n < 6 - \frac{4n-1}{n}$
(2) $b_n > \frac{4n-1}{2n}$
(3) The sequence $\langle b_n \rangle$ may diverge
(4) $a_{10000} < 6.1$
(5) $a_{10000} > 5.9$

Suppose two sequences $\langle a_n \rangle$ and $\langle b_n \rangle$ satisfy $b_n + \frac{4n-1}{n} < a_n < 3b_n$ for all positive integers $n$. Given that $\lim_{n \to \infty} a_n = 6$, select the correct options.\\
(1) $b_n < 6 - \frac{4n-1}{n}$\\
(2) $b_n > \frac{4n-1}{2n}$\\
(3) The sequence $\langle b_n \rangle$ may diverge\\
(4) $a_{10000} < 6.1$\\
(5) $a_{10000} > 5.9$

Paper Questions

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