设 $w_r=r(r+1)(r+2)\ldots(r+9)$。

(a) 证明

(b) 已知 $u_r=(r+1)(r+2)\ldots(r+9)$，求 $\displaystyle\sum_{r=1}^{n}u_r$，以 $n$ 表示。

(c) 已知 $v_r=x^{w_{r+1}}-x^{w_r}$，求使无穷级数

收敛的 $x$ 的取值集合，并在无穷和存在时给出该无穷和。

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。

两个乘积有九个相同因子，提出公因子后只剩首尾因子的差。

We have

Thus

as required.

由(a)把u写成w的相邻差，求和后仅留下最后一项与第一项。

By (a), $u_r=(w_{r+1}-w_r)/10$. For $n\ge6$, expanding the first and last three differences gives the cancellation below; for $1\le n\le5$, direct cancellation gives the same result. Hence

Since $w_1=10!$ and $w_{n+1}=(n+1)(n+2)\cdots(n+10)$,

同样先求部分和。收敛区间的两个端点必须另查，因为所有指数都是偶数，x为正负1时每一项实际为零。

For $n\ge6$, the partial sums telescope as follows. For $1\le n\le5$, directly cancelling the corresponding terms gives the same result:

If $|x|<1$, $w_{n+1}\to\infty$ implies $x^{w_{n+1}}\to0$, giving sum $-x^{10!}$.

Each $w_r$ is a product of ten consecutive positive integers and is even. Thus at $x=1$ or $x=-1$, every $x^{w_r}=1$, so every $v_r=0$ and the sum is $0$.

If $|x|>1$, the even exponents give $x^{w_{n+1}}=|x|^{w_{n+1}}\to\infty$, so the partial sums diverge.

Therefore the series converges exactly for $\boxed{-1\le x\le1}$, with sum