We are given a sequence $(u_n)$ defined by the recurrence relation $u_n = -\frac{1}{3}u_{n-1} + 1$ with initial value $u_0 = 3$. The problem consists of several parts: a) Calculate $u_2$, $u_3$ and $u_4$. b) Show that the sequence $(u_n)$ is neither arithmetic nor geometric. 2a) Define a new sequence $w_n = u_n - \frac{3}{4}$. Calculate $w_0$, $w_1$ and $w_2$. b) Show that the sequence $(w_n)$ is geometric and justify that it is convergent, specifying its limit. c) Deduce the limit of $(u_n)$. 3a) Express $w_n$ and $u_n$ as functions of $n$. Justify why $\lim_{n \to \infty} (-\frac{1}{3})^n = 0$. b) Define $S_n = w_1 + w_2 + \dots + w_n$. Calculate $S_n$ as a function of $n$, and then $\lim_{n \to \infty} S_n$. c) Define $T_n = u_1 + u_2 + \dots + u_n$. Deduce $T_n$ as a function of $n$ and then calculate $\lim_{n \to \infty} T_n$.

1. Problem Description

We are given a sequence $(u_n)$ defined by the recurrence relation $u_n = -\frac{1}{3}u_{n-1} + 1$ with initial value $u_0 = 3$. The problem consists of several parts:

a) Calculate $u_2$, $u_3$ and $u_4$.

b) Show that the sequence $(u_n)$ is neither arithmetic nor geometric.

2a) Define a new sequence $w_n = u_n - \frac{3}{4}$. Calculate $w_0$, $w_1$ and $w_2$.

b) Show that the sequence $(w_n)$ is geometric and justify that it is convergent, specifying its limit.

c) Deduce the limit of $(u_n)$.

3a) Express $w_n$ and $u_n$ as functions of $n$. Justify why $\lim_{n \to \infty} (-\frac{1}{3})^n = 0$.

b) Define $S_n = w_1 + w_2 + \dots + w_n$. Calculate $S_n$ as a function of $n$, and then $\lim_{n \to \infty} S_n$.

c) Define $T_n = u_1 + u_2 + \dots + u_n$. Deduce $T_n$ as a function of $n$ and then calculate $\lim_{n \to \infty} T_n$.

2. Solution Steps

1a) Calculate $u_2$, $u_3$ and $u_4$.

We have $u_0 = 3$ and $u_n = -\frac{1}{3}u_{n-1} + 1$.

$u_1 = -\frac{1}{3}u_0 + 1 = -\frac{1}{3}(3) + 1 = -1 + 1 = 0$.

$u_2 = -\frac{1}{3}u_1 + 1 = -\frac{1}{3}(0) + 1 = 1$.

$u_3 = -\frac{1}{3}u_2 + 1 = -\frac{1}{3}(1) + 1 = -\frac{1}{3} + 1 = \frac{2}{3}$.

$u_4 = -\frac{1}{3}u_3 + 1 = -\frac{1}{3}(\frac{2}{3}) + 1 = -\frac{2}{9} + 1 = \frac{7}{9}$.

b) Show that $(u_n)$ is neither arithmetic nor geometric.

For $(u_n)$ to be arithmetic, $u_1 - u_0 = u_2 - u_1$ must hold.

$u_1 - u_0 = 0 - 3 = -3$.

$u_2 - u_1 = 1 - 0 = 1$.

Since $-3 \ne 1$, $(u_n)$ is not arithmetic.

For $(u_n)$ to be geometric, $\frac{u_1}{u_0} = \frac{u_2}{u_1}$ must hold.

$\frac{u_1}{u_0} = \frac{0}{3} = 0$.

$\frac{u_2}{u_1} = \frac{1}{0}$, which is undefined.

Since $u_1 = 0$, $(u_n)$ is not geometric.

2a) On pose $w_n = u_n - \frac{3}{4}$. Calculate $w_0$, $w_1$ and $w_2$.

$w_0 = u_0 - \frac{3}{4} = 3 - \frac{3}{4} = \frac{12}{4} - \frac{3}{4} = \frac{9}{4}$.

$w_1 = u_1 - \frac{3}{4} = 0 - \frac{3}{4} = -\frac{3}{4}$.

$w_2 = u_2 - \frac{3}{4} = 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4}$.

b) Show that $(w_n)$ is geometric and justify that it is convergent.

$w_n = u_n - \frac{3}{4}$. Then $u_n = w_n + \frac{3}{4}$.

$w_n = u_n - \frac{3}{4} = -\frac{1}{3}u_{n-1} + 1 - \frac{3}{4} = -\frac{1}{3}(w_{n-1} + \frac{3}{4}) + \frac{1}{4} = -\frac{1}{3}w_{n-1} - \frac{1}{4} + \frac{1}{4} = -\frac{1}{3}w_{n-1}$.

Thus, $w_n = -\frac{1}{3}w_{n-1}$. Therefore, $(w_n)$ is a geometric sequence with common ratio $r = -\frac{1}{3}$ and first term $w_0 = \frac{9}{4}$.

Since $|r| = |-\frac{1}{3}| = \frac{1}{3} < 1$, the geometric sequence $(w_n)$ is convergent, and its limit is

0. $\lim_{n \to \infty} w_n = 0$.

c) Deduce the limit of $(u_n)$.

Since $w_n = u_n - \frac{3}{4}$, we have $u_n = w_n + \frac{3}{4}$.

Taking the limit as $n \to \infty$, we have $\lim_{n \to \infty} u_n = \lim_{n \to \infty} (w_n + \frac{3}{4}) = \lim_{n \to \infty} w_n + \frac{3}{4} = 0 + \frac{3}{4} = \frac{3}{4}$.

3a) Express $w_n$ and $u_n$ as functions of $n$.

Since $(w_n)$ is geometric with $w_0 = \frac{9}{4}$ and $r = -\frac{1}{3}$, we have $w_n = w_0 \cdot r^n = \frac{9}{4}(-\frac{1}{3})^n$.

Since $u_n = w_n + \frac{3}{4}$, we have $u_n = \frac{9}{4}(-\frac{1}{3})^n + \frac{3}{4}$.

Since $|\frac{-1}{3}| < 1$, $\lim_{n \to \infty} (-\frac{1}{3})^n = 0$.

b) Calculate $S_n = w_1 + w_2 + \dots + w_n$ and $\lim_{n \to \infty} S_n$.

$S_n$ is the sum of the first $n$ terms of the geometric sequence $(w_n)$ starting from $w_1$.

$S_n = \sum_{i=1}^{n} w_i = \sum_{i=1}^{n} \frac{9}{4}(-\frac{1}{3})^i = \frac{9}{4} \sum_{i=1}^{n} (-\frac{1}{3})^i$.

$\sum_{i=1}^{n} r^i = \frac{r(1-r^n)}{1-r}$

So, $S_n = \frac{9}{4} \frac{(-\frac{1}{3})(1 - (-\frac{1}{3})^n)}{1 - (-\frac{1}{3})} = \frac{9}{4} \frac{(-\frac{1}{3})(1 - (-\frac{1}{3})^n)}{\frac{4}{3}} = \frac{9}{4} (-\frac{1}{3}) (\frac{3}{4}) (1 - (-\frac{1}{3})^n) = -\frac{9}{16} (1 - (-\frac{1}{3})^n)$.

Then, $\lim_{n \to \infty} S_n = \lim_{n \to \infty} (-\frac{9}{16} (1 - (-\frac{1}{3})^n)) = -\frac{9}{16} (1 - 0) = -\frac{9}{16}$.

c) Calculate $T_n = u_1 + u_2 + \dots + u_n$ and $\lim_{n \to \infty} T_n$.

$T_n = \sum_{i=1}^n u_i = \sum_{i=1}^n (w_i + \frac{3}{4}) = \sum_{i=1}^n w_i + \sum_{i=1}^n \frac{3}{4} = S_n + n(\frac{3}{4}) = -\frac{9}{16} (1 - (-\frac{1}{3})^n) + \frac{3n}{4}$.

$\lim_{n \to \infty} T_n = \lim_{n \to \infty} (-\frac{9}{16} (1 - (-\frac{1}{3})^n) + \frac{3n}{4}) = \lim_{n \to \infty} (-\frac{9}{16} + \frac{3n}{4})$.

Since $\lim_{n \to \infty} \frac{3n}{4} = \infty$, we have $\lim_{n \to \infty} T_n = \infty$.

3. Final Answer

1a) $u_2 = 1$, $u_3 = \frac{2}{3}$, $u_4 = \frac{7}{9}$.

b) $(u_n)$ is neither arithmetic nor geometric.

2a) $w_0 = \frac{9}{4}$, $w_1 = -\frac{3}{4}$, $w_2 = \frac{1}{4}$.

b) $(w_n)$ is geometric with $r = -\frac{1}{3}$. $\lim_{n \to \infty} w_n = 0$.

c) $\lim_{n \to \infty} u_n = \frac{3}{4}$.

3a) $w_n = \frac{9}{4}(-\frac{1}{3})^n$, $u_n = \frac{9}{4}(-\frac{1}{3})^n + \frac{3}{4}$. $\lim_{n \to \infty} (-\frac{1}{3})^n = 0$.

b) $S_n = -\frac{9}{16} (1 - (-\frac{1}{3})^n)$. $\lim_{n \to \infty} S_n = -\frac{9}{16}$.

c) $T_n = -\frac{9}{16} (1 - (-\frac{1}{3})^n) + \frac{3n}{4}$. $\lim_{n \to \infty} T_n = \infty$.