用数学归纳法证明，对于所有正整数 $n$ 且 $x \in \mathbb{R}$，$x \ne 0, x \ne 1$， $\dfrac{1}{x-1}-\dfrac{1}{x}-\dfrac{1}{x^2}-\dots-\dfrac{1}{x^n}=\dfrac{1}{x^n(x-1)}.$ 已知 $u_n=\dfrac{1}{x-1}-\dfrac{1}{x}-\dfrac{1}{x^2}-\dots-\dfrac{1}{x^n},$ 求使 $\displaystyle\sum_{n=1}^{\infty}u_n$ 存在的 $x$ 的取值范围。

1. Establish the base case for n = 1.

Let $P_n$ be the statement: $\dfrac{1}{x-1}-\dfrac{1}{x}-\dfrac{1}{x^2}-\dots-\dfrac{1}{x^n}=\dfrac{1}{x^n(x-1)}$.

When $n=1$,
$\text{LHS} = \dfrac{1}{x-1}-\dfrac{1}{x} = \dfrac{x-(x-1)}{x(x-1)} = \dfrac{1}{x(x-1)}$
$\text{RHS} = \dfrac{1}{x^1(x-1)} = \dfrac{1}{x(x-1)}$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. State the inductive hypothesis.

Assume that $P_k$ is true for some positive integer $k$, i.e.,
$\dfrac{1}{x-1}-\dfrac{1}{x}-\dfrac{1}{x^2}-\dots-\dfrac{1}{x^k}=\dfrac{1}{x^k(x-1)}$

3. Prove the statement for n = k + 1 using the inductive hypothesis.

Consider $n=k+1$:
$\text{LHS} = \left(\dfrac{1}{x-1}-\dfrac{1}{x}-\dots-\dfrac{1}{x^k}\right)-\dfrac{1}{x^{k+1}}$

Substituting the assumption $P_k$:
$\text{LHS} = \dfrac{1}{x^k(x-1)} - \dfrac{1}{x^{k+1}}$
Expressing both terms with a common denominator $x^{k+1}(x-1)$:
$\text{LHS} = \dfrac{x}{x^{k+1}(x-1)} - \dfrac{x-1}{x^{k+1}(x-1)}$
$\text{LHS} = \dfrac{x - (x - 1)}{x^{k+1}(x-1)}$
$\text{LHS} = \dfrac{1}{x^{k+1}(x-1)}$

Since this is exactly the RHS expression evaluated at $n=k+1$, $P_{k+1}$ is true if $P_k$ is true.

4. State the formal conclusion for mathematical induction.

Since $P_1$ is true, and $P_{k+1}$ is true whenever $P_k$ is true, by the Principle of Mathematical Induction, $P_n$ is true for all positive integers $n$.

1. Use the result from part (i) to simplify the expression for $u_n$.

From part (i), we can substitute the simplified expression for $u_n$:
$u_n = \dfrac{1}{x^n(x-1)}$

2. Identify the infinite series as a geometric series and state the condition for its convergence.

The series $\displaystyle\sum_{n=1}^{\infty} u_n = \sum_{n=1}^{\infty} \dfrac{1}{x-1}\left(\dfrac{1}{x}\right)^n$ is a geometric series with common ratio $r = \dfrac{1}{x}$.

For the sum to infinity to exist, we require $|r| < 1$, so:
$\left|\dfrac{1}{x}\right| < 1$

3. Solve the inequality to find the range of values for $x$.

Since $\left|\dfrac{1}{x}\right| = \dfrac{1}{|x|}$, we have:
$\dfrac{1}{|x|} < 1$
$|x| > 1$
$\therefore x < -1 \text{ or } x > 1$