正数数列 $u_1$、$u_2$、$u_3$、$\ldots$ 满足 $u_1>1$，且当 $n\geqslant1$ 时，

(a) 通过考虑 $u_{n+1}-1$，或用其他方法，利用数学归纳法证明：对所有正整数 $n$，都有 $u_n>1$。

(b) 证明当 $n\geqslant1$ 时，$u_{n+1}<u_n$。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

按建议考察下一项减1，因式分解后用归纳假设判断每个因子和分母的符号。

The statement is true for $n=1$, since $u_1>1$ is given.

Assume $u_k>1$ for some positive integer $k$. Then

The inequality follows because $u_k-1$, $u_k+2$ and $u_k+4$ are all positive by the inductive hypothesis. Hence $u_{k+1}>1$.

Therefore, by mathematical induction, $\boxed{u_n>1}$ for every positive integer $n$.

相邻项作差并通分，再使用(a)已证明的下界。

For every $n\ge1$,

By (a), $u_n>1$, so the numerator is negative and the denominator positive. Thus $u_{n+1}-u_n<0$, giving $\boxed{u_{n+1}<u_n}$ for all $n\ge1$.