The problem defines a sequence $(u_n)$ with the initial term $u_0 = 1$ and the recursive formula $u_{n+1} = \frac{u_n}{2u_n+1}$ for all $n \in \mathbb{N}$. We are asked to: 1. Calculate $u_1, u_2, u_3$, and $u_4$.

1. Problem Description

The problem defines a sequence $(u_n)$ with the initial term $u_0 = 1$ and the recursive formula $u_{n+1} = \frac{u_n}{2u_n+1}$ for all $n \in \mathbb{N}$. We are asked to:

1. Calculate $u_1, u_2, u_3$, and $u_4$.

2. Define a new sequence $v_n = \frac{1}{u_n}$.

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

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

c. Calculate the limits of $u_n$ and $v_n$.

2. Solution Steps

1. Calculate $u_1, u_2, u_3, u_4$:

$u_0 = 1$

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

$u_2 = \frac{u_1}{2u_1+1} = \frac{\frac{1}{3}}{2(\frac{1}{3})+1} = \frac{\frac{1}{3}}{\frac{2}{3}+1} = \frac{\frac{1}{3}}{\frac{5}{3}} = \frac{1}{5}$

$u_3 = \frac{u_2}{2u_2+1} = \frac{\frac{1}{5}}{2(\frac{1}{5})+1} = \frac{\frac{1}{5}}{\frac{2}{5}+1} = \frac{\frac{1}{5}}{\frac{7}{5}} = \frac{1}{7}$

$u_4 = \frac{u_3}{2u_3+1} = \frac{\frac{1}{7}}{2(\frac{1}{7})+1} = \frac{\frac{1}{7}}{\frac{2}{7}+1} = \frac{\frac{1}{7}}{\frac{9}{7}} = \frac{1}{9}$

2. a. Show that $(v_n)$ is arithmetic:

$v_n = \frac{1}{u_n}$

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

$v_{n+1} - v_n = 2$

Since the difference between consecutive terms is constant, $(v_n)$ is an arithmetic sequence with common difference $d = 2$.

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

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

The general formula for an arithmetic sequence is:

$v_n = v_0 + nd$

$v_n = 1 + n(2) = 1 + 2n$

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

$u_n = \frac{1}{1+2n}$

c. Calculate the limits of $u_n$ and $v_n$:

$\lim_{n \to \infty} v_n = \lim_{n \to \infty} (1+2n) = \infty$

$\lim_{n \to \infty} u_n = \lim_{n \to \infty} \frac{1}{1+2n} = 0$

3. Final Answer

1. $u_1 = \frac{1}{3}, u_2 = \frac{1}{5}, u_3 = \frac{1}{7}, u_4 = \frac{1}{9}$

2. a. $(v_n)$ is an arithmetic sequence with first term $v_0 = 1$ and common difference $d=2$.

b. $v_n = 1+2n$, $u_n = \frac{1}{1+2n}$

c. $\lim_{n \to \infty} u_n = 0$, $\lim_{n \to \infty} v_n = \infty$