We are given a sequence $(u_n)$ defined by $u_0 = 1$ and $u_{n+1} = \frac{2u_n}{u_n+2}$ for all $n \in \mathbb{N}$. We define another sequence $(v_n)$ by $v_n = \frac{1}{u_n}$ for all $n \in \mathbb{N}$. We are asked to: 1. Calculate $u_1$, $u_2$, and $u_3$.

1. Problem Description

We are given a sequence $(u_n)$ defined by $u_0 = 1$ and $u_{n+1} = \frac{2u_n}{u_n+2}$ for all $n \in \mathbb{N}$.

We define another sequence $(v_n)$ by $v_n = \frac{1}{u_n}$ for all $n \in \mathbb{N}$.

We are asked to:

1. Calculate $u_1$, $u_2$, and $u_3$.

2. a) Show that $(v_n)$ is an arithmetic sequence and find its common difference and first term.

b) Express $v_n$ and $u_n$ as functions of $n$.

c) Study the convergence of $v_n$ and $u_n$.

3. Calculate $S_n = v_0 + v_1 + \dots + v_n$.

2. Solution Steps

1) Calculate $u_1$, $u_2$, and $u_3$.

$u_1 = \frac{2u_0}{u_0+2} = \frac{2(1)}{1+2} = \frac{2}{3}$.

$u_2 = \frac{2u_1}{u_1+2} = \frac{2(\frac{2}{3})}{\frac{2}{3}+2} = \frac{\frac{4}{3}}{\frac{8}{3}} = \frac{4}{8} = \frac{1}{2}$.

$u_3 = \frac{2u_2}{u_2+2} = \frac{2(\frac{1}{2})}{\frac{1}{2}+2} = \frac{1}{\frac{5}{2}} = \frac{2}{5}$.

2) a) Show that $(v_n)$ is an arithmetic sequence.

$v_n = \frac{1}{u_n}$, so $v_{n+1} = \frac{1}{u_{n+1}} = \frac{1}{\frac{2u_n}{u_n+2}} = \frac{u_n+2}{2u_n} = \frac{u_n}{2u_n} + \frac{2}{2u_n} = \frac{1}{2} + \frac{1}{u_n} = \frac{1}{2} + v_n$.

$v_{n+1} - v_n = \frac{1}{2}$. Therefore, $(v_n)$ is an arithmetic sequence with common difference $r = \frac{1}{2}$.

The first term is $v_0 = \frac{1}{u_0} = \frac{1}{1} = 1$.

b) Express $v_n$ and $u_n$ as functions of $n$.

Since $(v_n)$ is an arithmetic sequence with first term $v_0 = 1$ and common difference $r = \frac{1}{2}$, we have:

$v_n = v_0 + nr = 1 + \frac{n}{2} = \frac{2+n}{2}$.

Since $u_n = \frac{1}{v_n}$, we have $u_n = \frac{1}{\frac{2+n}{2}} = \frac{2}{2+n}$.

c) Study the convergence of $v_n$ and $u_n$.

$v_n = \frac{2+n}{2}$. As $n \to \infty$, $v_n \to \infty$. Therefore, $v_n$ diverges to infinity.

$u_n = \frac{2}{2+n}$. As $n \to \infty$, $u_n \to 0$. Therefore, $u_n$ converges to

0.

3) Calculate $S_n = v_0 + v_1 + \dots + v_n$.

Since $(v_n)$ is an arithmetic sequence, the sum of the first $n+1$ terms is:

$S_n = \sum_{k=0}^{n} v_k = \frac{(n+1)(v_0 + v_n)}{2} = \frac{(n+1)(1 + \frac{2+n}{2})}{2} = \frac{(n+1)(\frac{2+2+n}{2})}{2} = \frac{(n+1)(4+n)}{4} = \frac{n^2+5n+4}{4}$.

3. Final Answer

1) $u_1 = \frac{2}{3}$, $u_2 = \frac{1}{2}$, $u_3 = \frac{2}{5}$.

2) a) $(v_n)$ is an arithmetic sequence with first term $v_0 = 1$ and common difference $r = \frac{1}{2}$.

b) $v_n = \frac{2+n}{2}$, $u_n = \frac{2}{2+n}$.

c) $v_n$ diverges to $\infty$, $u_n$ converges to $0$.

3) $S_n = \frac{n^2+5n+4}{4}$.