The sequence un is defined by u1 = 2 and un+1 = (1)/(2)(un + (2)/(un)) for n >= 1. 1. Show that un > sqrt(2) for all n >= 1. 2. Show that un+1 < un for all n >= 1. 3. Hence explain why un converges, and find n infinity un. 4. Find the sum sumn=1^infinity (un - un+1).

We use the AM-GM inequality, which states that for non-negative numbers $a$ and $b$, $\frac{a+b}{2} \geq \sqrt{ab}$.
Let $a = u_n$ and $b = \frac{2}{u_n}$.
Since $u_1 = 2 > 0$, and the recurrence relation involves division by $u_n$, all $u_n$ must be positive.
Thus, $u_n > 0$ and $\frac{2}{u_n} > 0$.
Applying AM-GM to $u_n$ and $\frac{2}{u_n}$:

$$u_{n+1} = \frac{1}{2}\left(u_n + \frac{2}{u_n}\right) \geq \sqrt{u_n \cdot \frac{2}{u_n}}$$ $$u_{n+1} \geq \sqrt{2}$$

The equality holds if and only if $u_n = \frac{2}{u_n}$, which means $u_n^2 = 2$, so $u_n = \sqrt{2}$.
However, $u_1 = 2$, which is not equal to $\sqrt{2}$.
If $u_n > \sqrt{2}$, then $u_n \neq \frac{2}{u_n}$, so the inequality is strict: $u_{n+1} > \sqrt{2}$.
Since $u_1 = 2 > \sqrt{2}$, it follows that $u_2 > \sqrt{2}$, $u_3 > \sqrt{2}$, and so on.
Therefore, $u_n > \sqrt{2}$ for all $n \geq 1$.

To show that $u_{n+1} < u_n$, we need to show that $u_n - u_{n+1} > 0$.

$$u_n - u_{n+1} = u_n - \frac{1}{2}\left(u_n + \frac{2}{u_n}\right)$$ $$u_n - u_{n+1} = u_n - \frac{u_n}{2} - \frac{1}{u_n}$$ $$u_n - u_{n+1} = \frac{u_n}{2} - \frac{1}{u_n}$$ $$u_n - u_{n+1} = \frac{u_n^2 - 2}{2u_n}$$

From Step 1, we established that $u_n > \sqrt{2}$ for all $n \geq 1$.
This implies $u_n^2 > (\sqrt{2})^2 = 2$, so $u_n^2 - 2 > 0$.
Also, since $u_n > \sqrt{2}$, it means $u_n$ is positive, so $2u_n > 0$.
Therefore, $\frac{u_n^2 - 2}{2u_n} > 0$.
This shows that $u_n - u_{n+1} > 0$, which means $u_{n+1} < u_n$ for all $n \geq 1$.
The sequence $u_n$ is strictly decreasing.

From Step 1, the sequence $u_n$ is bounded below by $\sqrt{2}$ (i.e., $u_n > \sqrt{2}$ for all $n \geq 1$).
From Step 2, the sequence $u_n$ is monotonically decreasing (i.e., $u_{n+1} < u_n$ for all $n \geq 1$).
A fundamental theorem in real analysis states that any sequence that is both monotonic and bounded converges. Therefore, $u_n$ converges.

To find the limit, let $L = \lim_{n \to \infty} u_n$.
If $u_n$ converges to $L$, then $u_{n+1}$ also converges to $L$.
Taking the limit of the recurrence relation:

$$\lim_{n \to \infty} u_{n+1} = \lim_{n \to \infty} \frac{1}{2}\left(u_n + \frac{2}{u_n}\right)$$ $$L = \frac{1}{2}\left(L + \frac{2}{L}\right)$$

Multiply by 2:

$$2L = L + \frac{2}{L}$$

Subtract $L$ from both sides:

$$L = \frac{2}{L}$$

Multiply by $L$:

$$L^2 = 2$$ $$L = \pm \sqrt{2}$$

Since we established in Step 1 that $u_n > \sqrt{2}$ for all $n \geq 1$, the limit $L$ must be positive.
Therefore, the limit is $\boxed{\sqrt{2}}$.

This is a telescoping series. Let $S_N$ be the $N$-th partial sum:

$$S_N = \sum_{n=1}^{N} \left(u_n - u_{n+1}\right)$$ $$S_N = (u_1 - u_2) + (u_2 - u_3) + (u_3 - u_4) + \dots + (u_N - u_{N+1})$$

All intermediate terms cancel out, leaving:

$$S_N = u_1 - u_{N+1}$$

The sum of the infinite series is the limit of the partial sums as $N \to \infty$:

$$\sum_{n=1}^{\infty} \left(u_n - u_{n+1}\right) = \lim_{N \to \infty} S_N = \lim_{N \to \infty} (u_1 - u_{N+1})$$

We know $u_1 = 2$.
From Step 3, we found that $\lim_{N \to \infty} u_{N+1} = \sqrt{2}$.
Substituting these values:

$$\sum_{n=1}^{\infty} \left(u_n - u_{n+1}\right) = 2 - \sqrt{2}$$

The sum is $\boxed{2 - \sqrt{2}}$.