Let $ a\geq 2$ be a real number; with the roots $ x_{1}$ and $ x_{2}$ of the equation $ x^2\minus{}ax\plus{}1\equal{}0$ we build the sequence with $ S_{n}\equal{}x_{1}^n \plus{} x_{2}^n$.
[b]a)[/b]Prove that the sequence $ \frac{S_{n}}{S_{n\plus{}1}}$, where $ n$ takes value from $ 1$ up to infinity, is strictly non increasing.
[b]b)[/b]Find all value of $ a$ for the which this inequality hold for all natural values of $ n$ $ \frac{S_{1}}{S_{2}}\plus{}\cdots \plus{}\frac{S_{n}}{S_{n\plus{}1}}>n\minus{}1$
**Part (a): Prove that the sequence $\frac{S_n}{S_{n+1}}$, where $n$ takes values from $1$ up to infinity, is strictly non-increasing.**

1. Given the quadratic equation \(x^2 - ax + 1 = 0\), the roots \(x_1\) and \(x_2\) are given by:
   \[
   x_1, x_2 = \frac{a \pm \sqrt{a^2 - 4}}{2}
   \]
   Since \(a \geq 2\), the discriminant \(a^2 - 4 \geq 0\), ensuring that \(x_1\) and \(x_2\) are real numbers.

2. The sequence \(S_n = x_1^n + x_2^n\) satisfies the recurrence relation:
   \[
   S_{n+2} = aS_{n+1} - S_n
   \]
   This can be derived by noting that \(x_1\) and \(x_2\) satisfy the original quadratic equation:
   \[
   x_1^2 = ax_1 - 1 \quad \text{and} \quad x_2^2 = ax_2 - 1
   \]
   Therefore,
   \[
   x_1^{n+2} = ax_1^{n+1} - x_1^n \quad \text{and} \quad x_2^{n+2} = ax_2^{n+1} - x_2^n
   \]
   Adding these, we get:
   \[
   S_{n+2} = aS_{n+1} - S_n
   \]

3. To show that \(\frac{S_n}{S_{n+1}}\) is strictly non-increasing, consider:
   \[
   S_{n+1}^2 - S_n S_{n+2} = (x_1^{n+1} + x_2^{n+1})^2 - (x_1^n + x_2^n)(x_1^{n+2} + x_2^{n+2})
   \]
   Using the recurrence relation:
   \[
   S_{n+2} = aS_{n+1} - S_n
   \]
   we get:
   \[
   S_{n+1}^2 - S_n S_{n+2} = S_{n+1}^2 - S_n(aS_{n+1} - S_n) = S_n^2 - aS_n S_{n+1} + S_{n+1}^2
   \]
   This simplifies to:
   \[
   S_{n+1}^2 - S_n S_{n+2} = -x_1^n x_2^n (x_1 - x_2)^2 \leq 0
   \]
   Since \(S_n > 0\), it follows that:
   \[
   \frac{S_n}{S_{n+1}} \geq \frac{S_{n+1}}{S_{n+2}}
   \]
   Hence, the sequence \(\frac{S_n}{S_{n+1}}\) is strictly non-increasing.

**Part (b): Find all values of \(a\) for which the inequality \(\frac{S_1}{S_2} + \cdots + \frac{S_n}{S_{n+1}} > n - 1\) holds for all natural values of \(n\).**

1. From part (a), we have:
   \[
   \frac{S_n}{S_{n+1}} \geq \frac{S_{n+1}}{S_{n+2}}
   \]
   This implies:
   \[
   \left(\frac{S_n}{S_{n+1}}\right)^2 - a\frac{S_n}{S_{n+1}} + 1 \leq 0
   \]
   Solving the quadratic inequality:
   \[
   t^2 - at + 1 \leq 0
   \]
   where \(t = \frac{S_n}{S_{n+1}}\), we find the roots:
   \[
   t = \frac{a \pm \sqrt{a^2 - 4}}{2}
   \]
   Since \(\frac{S_n}{S_{n+1}} > 0\), we have:
   \[
   \frac{a - \sqrt{a^2 - 4}}{2} \leq \frac{S_n}{S_{n+1}} \leq \frac{a + \sqrt{a^2 - 4}}{2}
   \]

2. Summing the inequalities:
   \[
   \sum_{k=1}^n \frac{S_k}{S_{k+1}} \geq n \cdot \frac{a - \sqrt{a^2 - 4}}{2}
   \]
   We need:
   \[
   n \cdot \frac{a - \sqrt{a^2 - 4}}{2} > n - 1
   \]
   Dividing both sides by \(n\):
   \[
   \frac{a - \sqrt{a^2 - 4}}{2} > 1 - \frac{1}{n}
   \]
   As \(n \to \infty\), the right-hand side approaches 1, so we need:
   \[
   \frac{a - \sqrt{a^2 - 4}}{2} > 1
   \]
   Solving for \(a\):
   \[
   a - \sqrt{a^2 - 4} > 2
   \]
   Squaring both sides:
   \[
   a^2 - a\sqrt{a^2 - 4} + 4 > 4
   \]
   Simplifying:
   \[
   a^2 - a\sqrt{a^2 - 4} > 0
   \]
   Since \(a \geq 2\), this inequality holds for all \(a \geq 2\).

The final answer is \( \boxed{ a \geq 2 } \)