# Prob. 10, Sec. 3.5, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: If $x_1 < x_2$ are arbitrary real numbers and . . .

Here is Prob. 10, Sec. 3.5, in the book Introduction to Real Analysis by Rovert G. Bartle and Donald R. Sherbert, 4th edition:

If $x_1 < x_2$ are arbitrary real numbers and $x_n \colon= \frac{1}{2} \left( x_{n-2} + x_{n-1} \right)$ for $n > 2$, show that $\left( x_n \right)$ is convergent. What is its limit?

My Attempt:

As $x_1 < x_2$, so $$x_1 < \frac{1}{2} \left( x_1 + x_2 \right) < x_2,$$ that is, $$x_1 < x_3 < x_2. \tag{1}$$ And also $$\left\lvert x_3 - x_2 \right\rvert = x_2 - x_3 = \frac{1}{2} \left( x_2 - x_1 \right). \tag{2}$$

Now as $x_3 < x_2$, so $$x_3 < \frac{1}{2} \left( x_3 + x_2 \right) < x_2,$$ that is, $$x_3 < x_4 < x_2. \tag{3}$$ And also $$\left\lvert x_4 - x_3 \right\rvert = x_4 - x_3 = \frac{1}{2} \left( x_2 - x_3 \right) = \frac{1}{2} \left\lvert x_2 - x_3 \right\rvert = \frac{1}{2^2} \left( x_2 - x_1 \right). \tag{4}$$

Now as $x_3 < x_4$, so $$x_3 < \frac{1}{2} \left( x_3 + x_4 \right) < x_4,$$ that is, $$x_3 < x_5 < x_4. \tag{5}$$ And also $$\left\lvert x_5 - x_4 \right\rvert = x_4 - x_5 = \frac{1}{2} \left( x_4 - x_3 \right) = \frac{1}{2} \left\lvert x_4 - x_3 \right\rvert = \frac{1}{2^3} \left( x_2 - x_1 \right). \tag{6}$$

Now as $x_5 < x_4$, so $$x_5 < \frac{1}{2} \left( x_5 + x_4 \right) < x_4,$$ that is, $$x_5 < x_6 < x_4. \tag{7}$$ And also $$\left\lvert x_6 - x_5 \right\rvert = x_6 - x_5 = \frac{1}{2} \left( x_4 - x_5 \right) = \frac{1}{2} \left\lvert x_5 - x_4 \right\rvert = \frac{1}{2^4} \left( x_2 - x_1 \right). \tag{8}$$

From (1), (3), (5), and (7), we get $$x_1 < x_3 < x_5 < x_6 < x_4 < x_2. \tag{9}$$

Now let $k \in \mathbb{N}$ such that $k \geq 3$, and suppose that $$x_1 < x_3 < x_5 < \cdots < x_{2k-1} < x_{2k} < x_{2k-2} < \cdots < x_4 < x_2. \tag{10}$$ And suppose also that $$\left\lvert x_{2k} - x_{2k-1} \right\vert = \frac{1}{ 2^{2k-2} } \left( x_2 - x_1 \right). \tag{11}$$

Then as $x_{2k-1} < x_{2k}$, so $$x_{2k-1} < \frac{1}{2} \left( x_{2k-1} + x_{2k} \right) < x_{2k},$$ that is, $$x_{2k-1} < x_{2k+1} < x_{2k }. \tag{12}$$ And also $$\left\lvert x_{2k+1} - x_{2k} \right\rvert = x_{2k} - x_{2k+1} = \frac{1}{2} \left( x_{2k} - x_{2k-1} \right) = \frac{1}{2} \left\lvert x_{2k} - x_{2k-1} \right\rvert = \frac{1}{ 2^{2k-1} } \left( x_2 - x_1 \right). \tag{13}$$

Now as $x_{2k+1} < x_{2k}$, so $$x_{2k+1} < \frac{1}{2} \left( x_{2k+1} + x_{2k} \right) < x_{2k},$$ that is, $$x_{2k+1} < x_{2k+2} < x_{2k}. \tag{14}$$

From (10), (12), and (14), we can conclude that $$x_1 < x_3 < x_5 < \cdots < x_{2k-1} < x_{2k+1} < x_{2k+2} < x_{2k} < \cdots < x_6 < x_4 < x_2 \tag{15}$$ for all $k \in \mathbb{N}$. [I'm not really sure what to do with this relation though.]

Also from (11) and (13), we can conclude that $$\left\lvert x_{n+1} - x_n \right\rvert = \frac{1}{2^{n-1} } \left( x_2 - x_1 \right). \tag{A}$$ for all $n \in \mathbb{N}$ such that $n \geq 2$.

So for any natural numbers $m$ and $n$ such that $m > n$, we obtain \begin{align} \left\lvert x_m - x_n \right\rvert &\leq \left\lvert x_m - x_{m-1} \right\rvert + \cdots + \left\lvert x_{n+1} - x_n \right\rvert \\ &= \left( \frac{1}{2^{m-2} } + \cdots + \frac{1}{2^{n-1} } \right) \left( x_2 - x_1 \right) \\ &= \left( \frac{1}{2^{n-1} } + \frac{1}{2^{n} } \cdots + \frac{1}{2^{m-2} } \right) \left( x_2 - x_1 \right) \\ &= \frac{1}{2^{n-1}} \left( 1 + \frac{1}{2} + \cdots + \frac{1}{2^{m-n-1}} \right) \left( x_2 - x_1 \right) \\ &= \frac{1}{2^{n-1}} \frac{ 1 - \frac{1}{2^{m-n}} }{ 1 - \frac{1}{2} } \left( x_2 - x_1 \right) \\ &= \frac{1}{2^{n-2}} \left( 1 - \frac{1}{2^{m-n}} \right) \left( x_2 - x_1 \right) \\ &< \frac{1}{2^{n-2}} \left( x_2 - x_1 \right). \tag{B} \end{align}

So, given $\varepsilon > 0$, if we choose a natural number $N$ so that $$N > \frac{ 4 \left(x_2 - x_1 \right) }{ \varepsilon },$$ then $$2^N > N > \frac{ 4 \left(x_2 - x_1 \right) }{ \varepsilon },$$ and so $$\varepsilon > \frac{ 4 \left( x_2 - x_1 \right) }{ 2^N}.$$ and then we see from (B) that, whenever $m$ and $n$ are any natural numbers such that $m > n > N$, we have $$\left\lvert x_m - x_n \right\rvert < \frac{1}{2^{n-2}} \left( x_2 - x_1 \right) = \frac{4 \left( x_2 - x_1 \right) }{ 2^n} < \frac{4 \left( x_2 - x_1 \right)}{2^N} < \varepsilon.$$

Thus our sequence is a Cauchy sequence of real numbers. Therefore, this sequence is convergent. Let $x$ be the limit of this sequence. Then the subsequences $\left( x_{2k-1} \right)_{k \in \mathbb{N}}$ and $\left( x_{2k } \right)_{k \in \mathbb{N}}$ also converge to this same limit $x$.

Now $$x_3 = \frac{x_2 + x_1 }{2},$$ and so $$x_4 = \frac{x_3 + x_2}{2} = \frac{ \frac{x_2 + x_1 }{2} + x_2 }{2} = \frac{ 3x_2 + x_1 }{4},$$ and hence $$x_5 = \frac{ x_4 + x_3 }{2} = \frac{ \frac{ 3x_2 + x_1 }{4} + \frac{x_2 + x_1}{2} }{2} = \frac{ 5 x_2 + 3 x_1 }{8}.$$
Thence $$x_6 = \frac{ x_5 + x_4 }{2} = \frac{ \frac{ 5 x_2 + 3 x_1 }{8} + \frac{ 3 x_2 + x_1 }{4} }{2} = \frac{ 11 x_2 + 5 x_1 }{16},$$ and so $$x_7 = \frac{ x_6 + x_5 }{2} = \frac{ \frac{ 11 x_2 + 5 x_1 }{16} + \frac{ 5 x_2 + 3 x_1 }{8} }{2} = \frac{ 21 x_2 + 11 x_1 }{32}.$$ Then $$x_8 = \frac{ x_7 + x_6 }{2} = \frac{ \frac{ 21 x_2 + 11 x_1 }{32} + \frac{ 11 x_2 + 5 x_1 }{ 16 } }{2} = \frac{ 43 x_2 + 21 x_1 }{ 64},$$ and so $$x_9 = \frac{ x_8 + x_7 }{2} = \frac{ \frac{ 43 x_2 + 21 x_1 }{64} + \frac{ 21 x_2 + 11 x_1 }{32} }{2} = \frac{ 85 x_2 + 43 x_1 }{128}.$$

Generalising from these formulas for $x_3$ through $x_9$, we suppose that $k \in \mathbb{N}$ such that $k > 1$ and that $$x_{2k - 1} = \frac{ \frac{ 2^{2k-2} - 1}{3} x_2 + \frac{ 2^{2k-3} + 1}{3} x_1 }{ 2^{2k-3} } = \frac{ \left( 2^{2k-2} - 1 \right) x_2 + \left( 2^{2k-3} + 1 \right) x_1 }{ 3 \times 2^{2k-3} },$$ and $$x_{2k} = \frac{ \frac{ 2^{2k-1} + 1}{3} x_2 + \frac{ 2^{2k-2} - 1}{3} x_1 }{2^{2k-2}} = \frac{ \left( 2^{2k-1} + 1 \right) x_2 + \left( 2^{2k-2} - 1 \right) x_1 }{ 3 \times 2^{2k-2} }.$$

Then we find that \begin{align} x_{2k+1} &= \frac{ x_{2k} + x_{2k-1} }{2} \\ &= \frac{ \frac{ \left( 2^{2k-1} + 1 \right) x_2 + \left( 2^{2k-2} - 1 \right) x_1 }{ 3 \times 2^{2k-2} } + \frac{ \left( 2^{2k-2} - 1 \right) x_2 + \left( 2^{2k-3} + 1 \right) x_1 }{ 3 \times 2^{2k-3} } }{2} \\ &= \frac{ \left( 2^{2k-1} + 1 \right) x_2 + \left( 2^{2k-2} - 1 \right) x_1 + 2 \left( 2^{2k-2} - 1 \right) x_2 + 2 \left( 2^{2k-3} + 1 \right) x_1 }{ 3 \times 2^{2k-1} } \\ &= \frac{ \left( 2^{2k} - 1 \right) x_2 + \left( 2^{2k-1} + 1 \right) x_1 }{ 3 \times 2^{2k-1} }, \end{align} and hence \begin{align} x_{2k+2} &= \frac{ x_{2k+1} + x_{2k} }{2} \\ &= \frac{ \frac{ \left( 2^{2k} - 1 \right) x_2 + \left( 2^{2k-1} + 1 \right) x_1 }{ 3 \times 2^{2k-1} } + \frac{ \left( 2^{2k-1} + 1 \right) x_2 + \left( 2^{2k-2} - 1 \right) x_1 }{ 3 \times 2^{2k-2} } }{2} \\ &= \frac{ \left( 2^{2k} - 1 \right) x_2 + \left( 2^{2k-1} + 1 \right) x_1 + 2 \left( 2^{2k-1} + 1 \right) x_2 + 2 \left( 2^{2k-2} - 1 \right) x_1 }{ 3 \times 2^{2k} } \\ &= \frac{ \left( 2^{2k+1} + 1 \right) x_2 + \left( 2^{2k} - 1 \right)x_1 }{ 3 \times 2^{2k} }. \end{align}

Thus the induction is complete, and so we can conclude that, for every natural number $k > 1$, we have $$x_{2k-1} = \frac{ \left( 2^{2k-2} - 1 \right) x_2 + \left( 2^{2k-3} + 1 \right) x_1 }{ 3 \times 2^{2k-3} } = \frac{1}{3} \left[ \left( 2 - \frac{1}{2^{2k-3}} \right) x_2 + \left( 1 + \frac{1}{2^{2k-3}} \right) x_1 \right], \tag{C}$$ and $$x_{2k} = \frac{ \left( 2^{2k-1} + 1 \right) x_2 + \left( 2^{2k-2} - 1 \right) x_1 }{ 3 \times 2^{2k-2} } = \frac{1}{3} \left[ \left( 2 + \frac{1}{2^{2k-2}} \right) x_2 + \left( 1 - \frac{1}{ 2^{2k-2} } \right) x_1 \right]. \tag{D}$$

Now as $$\lim_{k \to \infty} \frac{ 1 }{2^{2k-2} } = 0 = \lim_{k \to \infty} \frac{1}{2^{2k-3}},$$ so from formulas (C) and (D), we can conclude that $$\lim_{k \to \infty} x_{2k-1} = \frac{ 2x_2 + x_1 }{3} = \lim_{ k \to \infty} x_{2k}.$$ Therefore, $$\lim_{n \to \infty} x_n = \frac{ 2x_2 + x_1 }{3}.$$

Is what I've done correct? I know I have protracted my solution way way beyond the solution expected, but what is the shorter approach to it, I wonder?

Yes, your attempt is correct, and you made a mistake and I don't really want to find it. But it is very tedious. Here is another approach: First note that for $n\geq 3$ you get $$x_n = x_1 + \frac{4(x_2-x_1)}{3}\left( \left(-\frac{1}{2}\right)^n + \frac{1}{2} \right)$$ (look for the first terms of the sequence until you find the pattern, then prove it by induction). So, as the sequence $(a^n)$ converges to $0$ for every $a\in ]-1,1[$, you obtain that the sequence $(x_n)$ is convergent by elementary algebra of sequences and that.

$$x_n \to x_1 + \frac{4(x_2-x_1)}{3}\left( \frac{1}{2} \right) = \frac{x_1+2x_2}{3}.$$

As yo will see, this is exactly what you have done, but there is no need to prove the convergence of $(x_n)$ at first as this can be done from this closed expression for $x_n$.

This can be easily solved considering it as a difference equation

$$x_n -\frac{1}{2}x_{n-1}-\frac{1}{2}x_{n-2} = 0$$

Making $x_n = a^n$ and substituting

$$a^n-\frac{1}{2}a^{n-1}-\frac{1}{2}a^{n-2} = 0 \Rightarrow a^2-\frac{1}{2}a-\frac{1}{2} = 0 \ \ \ \hbox{this guarantees the difference recursion}$$

then the solution is

$$x_n = C_1 (-1) ^n 2^{-n}+1 \cdot C_2$$

now we can conclude about convergence

• @CarlosAjila I have corrected the mistake(s) that you've pointed out. – Saaqib Mahmood Apr 23 '18 at 21:21