Prove by induction that, for all positive integers $n$ and for $x\in\mathbb{R}$, $x\ne 0,\ x\ne 1$, $\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-\dots-\frac{1}{x^n}=\frac{1}{x^n(x-1)}\,.$ Find the range of values of $x$ for which $\sum_{n=1}^{\infty} u_n$ exists given that $u_n=\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-\dots-\frac{1}{x^n}\,.$

1. Define the statement and verify the base case for $n=1$.

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

For $n=1$:
$\text{LHS} = \frac{1}{x-1} - \frac{1}{x} = \frac{x - (x-1)}{x(x-1)} = \frac{1}{x(x-1)}$
$\text{RHS} = \frac{1}{x^1(x-1)} = \frac{1}{x(x-1)}$

Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. Formulate the inductive hypothesis.

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

We want to show that $P_{k+1}$ is also true.

3. Use the inductive hypothesis to prove the statement for $n=k+1$.

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

Substituting $P_k$:
$\text{LHS} = \frac{1}{x^k(x-1)} - \frac{1}{x^{k+1}}$

Making the denominators common by multiplying the numerator and denominator of the first term by $x$ and the second term by $(x-1)$:
$\text{LHS} = \frac{x}{x^{k+1}(x-1)} - \frac{x-1}{x^{k+1}(x-1)}$
$\text{LHS} = \frac{x - (x-1)}{x^{k+1}(x-1)}$
$\text{LHS} = \frac{1}{x^{k+1}(x-1)} = \text{RHS}$

Thus, $P_{k+1}$ is true if $P_k$ is true.

4. State the final conclusion for mathematical induction.

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

1. Express the series terms using the result from part (i) and identify the sequence type.

From part (i), we have $u_n = \frac{1}{x^n(x-1)}$.

The series is $\sum_{n=1}^{\infty} u_n = \sum_{n=1}^{\infty} \frac{1}{x^n(x-1)}$.
This is an infinite geometric series with first term $u_1 = \frac{1}{x(x-1)}$ and common ratio $r = \frac{1}{x}$.

2. State the condition for the infinite geometric series to converge.

For the sum to infinity of a geometric series to exist, the common ratio $r$ must satisfy $|r| < 1$.

Therefore, we require $\left| \frac{1}{x} \right| < 1$.

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

Since $x \neq 0$, we can rewrite the inequality as:
$\frac{1}{|x|} < 1$
$|x| > 1$

Solving this modulus inequality gives:
$x < -1 \text{ or } x > 1$