Given a sequence $(u_n)$ such that $u_0 = 3$ and $u_{n+1} = -\frac{1}{3}u_n + 1$. (a) Calculate $u_2$, $u_3$, and $u_4$. (b) Show that $(u_n)$ is neither arithmetic nor geometric. (2a) Let $w_n = u_n - \frac{3}{4}$. Calculate $w_0$, $w_1$, and $w_2$. (b) Show that $(w_n)$ is geometric and justify that it is convergent by specifying its limit. (c) Deduce the limit of $(u_n)$ from the previous results. (3a) Express $w_n$ and $u_n$ as a function of $n$. Justify why $\lim_{n\to\infty} (-\frac{1}{3})^n = 0$. (b) Let $S_n = w_1 + w_2 + \dots + w_n$. Calculate $S_n$ as a function of $n$, and then $\lim_{n\to\infty} S_n$. (c) Let $T_n = u_1 + u_2 + \dots + u_n$. Deduce $T_n$ as a function of $n$, and then $\lim_{n\to\infty} T_n$.

1. Problem Description

Given a sequence $(u_n)$ such that $u_0 = 3$ and $u_{n+1} = -\frac{1}{3}u_n + 1$.

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

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

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

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

(c) Deduce the limit of $(u_n)$ from the previous results.

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

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

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

2. Solution Steps

(1a)

$u_0 = 3$

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

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

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

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

(1b)

If $(u_n)$ is arithmetic, then $u_1 - u_0 = u_2 - u_1$. We have $0-3 = -3$ and $1-0 = 1$. Since $-3 \neq 1$, the sequence is not arithmetic.

If $(u_n)$ is geometric, then $\frac{u_1}{u_0} = \frac{u_2}{u_1}$. We have $\frac{0}{3} = 0$ and $\frac{1}{0}$ is undefined. Even if we ignore $u_1$, we can see $\frac{u_2}{u_1}$ is undefined, so the sequence is not geometric.

Alternatively, if $(u_n)$ is geometric, $\frac{u_1}{u_0} = \frac{u_2}{u_1}$. We have $\frac{u_2}{u_1} = \frac{1}{0}$ and $\frac{u_3}{u_2} = \frac{2/3}{1} = \frac{2}{3}$. Hence, since $\frac{1}{0} \neq \frac{2}{3}$, the sequence is not geometric.

(2a)

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

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

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

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

(2b)

We want to show that $\frac{w_{n+1}}{w_n}$ is constant.

$w_{n+1} = u_{n+1} - \frac{3}{4} = -\frac{1}{3}u_n + 1 - \frac{3}{4} = -\frac{1}{3}u_n + \frac{1}{4}$

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

Thus, $\frac{w_{n+1}}{w_n} = -\frac{1}{3}$. Therefore, $(w_n)$ is geometric with common ratio $r = -\frac{1}{3}$.

Since $|r| = |-\frac{1}{3}| = \frac{1}{3} < 1$, the sequence converges to

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

(2c)

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

Thus $\lim_{n\to\infty} u_n = \lim_{n\to\infty} (w_n + \frac{3}{4}) = \lim_{n\to\infty} w_n + \frac{3}{4} = 0 + \frac{3}{4} = \frac{3}{4}$.

(3a)

Since $(w_n)$ is a geometric sequence with $w_0 = \frac{9}{4}$ and ratio $r = -\frac{1}{3}$, we have

$w_n = w_0 r^n = \frac{9}{4} (-\frac{1}{3})^n$

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

$u_n = \frac{9}{4} (-\frac{1}{3})^n + \frac{3}{4}$

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

If $n$ is even, then $(-\frac{1}{3})^n = (\frac{1}{3})^n > 0$

If $n$ is odd, then $(-\frac{1}{3})^n = -(\frac{1}{3})^n < 0$

As $n \to \infty$, $(\frac{1}{3})^n \to 0$. Hence $\lim_{n\to\infty} (-\frac{1}{3})^n = 0$

(3b)

$S_n = w_1 + w_2 + \dots + w_n$. This is the sum of the first $n$ terms of a geometric sequence with first term $w_1 = -\frac{3}{4}$ and ratio $r = -\frac{1}{3}$.

$S_n = \frac{w_1(1-r^n)}{1-r} = \frac{-\frac{3}{4}(1-(-\frac{1}{3})^n)}{1-(-\frac{1}{3})} = \frac{-\frac{3}{4}(1-(-\frac{1}{3})^n)}{\frac{4}{3}} = -\frac{9}{16}(1-(-\frac{1}{3})^n)$

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

(3c)

$T_n = u_1 + u_2 + \dots + u_n$

$T_n = (w_1 + \frac{3}{4}) + (w_2 + \frac{3}{4}) + \dots + (w_n + \frac{3}{4})$

$T_n = (w_1 + w_2 + \dots + w_n) + n(\frac{3}{4})$

$T_n = S_n + \frac{3n}{4}$

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

Since $\lim_{n\to\infty} (-\frac{1}{3})^n = 0$, we have $\lim_{n\to\infty} [-\frac{9}{16}(1-(-\frac{1}{3})^n)] = -\frac{9}{16}$. However, $\lim_{n\to\infty} \frac{3n}{4} = \infty$.

Thus, $\lim_{n\to\infty} T_n = \infty$.

3. Final Answer

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

(1b) $(u_n)$ is neither arithmetic nor geometric.

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

(2b) $(w_n)$ is geometric with common ratio $r = -\frac{1}{3}$. $\lim_{n\to\infty} w_n = 0$

(2c) $\lim_{n\to\infty} u_n = \frac{3}{4}$

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

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

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