正数数列 $u_1$、$u_2$、$u_3$、$\ldots$ 满足 $u_1<5$，且当 $n\geqslant1$ 时，

(a) 通过考虑 $5-u_{n+1}$，利用数学归纳法证明：对所有正整数 $n$，都有 $u_n<5$。

(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。

按要求对5减下一项通分，用归纳假设保证分子和分母都为正。

The base case $u_1<5$ is given.

Assume $u_k<5$ for some positive integer $k$. Since the sequence is positive, $u_k+2>0$. Then

The numerator is positive by the inductive hypothesis, so $u_{k+1}<5$. Therefore, by mathematical induction, $\boxed{u_n<5}$ for every positive integer $n$.

相邻项作差并因式分解，利用正数条件和(a)得到的上界判断符号。

For $n\ge1$,

Here $5-u_n>0$ by (a), and $u_n+1,u_n+2>0$ because $u_n>0$. Hence $\boxed{u_{n+1}>u_n}$ for every $n\ge1$.