How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=\frac{a_{n-1}+a_{n-2}}{2}$ converges to $\frac23$?

Question

How to prove that the sequence defined by $a_1=0$, $a_2=1$, $a_n=\frac{a_{n-1}+a_{n-2}}{2}$ converges to $\frac23$?

If we analyse terms: $$ 0,1,\frac{1}{2},\frac{3}{4},\frac{5}{8},\cdots. $$

I'm asked to do this using a previously proved theorem which says that

if you have two sequences ${b_n}$ and ${c_n}$ converging both to the same limit $L$, then the sequence $a_n$ defined as $b_1,c_1,b_2,c_2,b_3,\dots$ converge to $L$.

In this case, $b_n$ would be $0,\cfrac{1}{2},\cfrac{5}{8},\cfrac{21}{32},\dots$ and $c_n$ would be $1,\cfrac{3}{4},\cfrac{11}{16},\cfrac{43}{64},\dots$

From here it's seems all I need to do is prove that $b_n$ and $c_n$ converge to $\cfrac{2}{3}$ and then, by the theorem, $a_n$ converges to $\cfrac{2}{3}$.

I need to define them because I need to prove $b_n$ and $c_n$ are monotone and bounded, in order to use the monotone convergence theorem. I've got troubles when trying to define $b_n$ and $c_n$. Can anyone help me to define them?

Answer 1

Here is an alternative (maybe easier) way.

By definition of the sequence one has $$ \begin{align} a_1+a_2&=2a_3\\ a_2+a_3&=2a_4\\ &\vdots\\ a_{n-1}+a_{n}&=2a_{n+1} \end{align} $$ Adding together these identities one gets $$ a_{n+1}=1-\frac12 a_{n},\quad n\geqslant 3.\tag{1} $$ Let $b_n=a_n-\frac23$. Then (1) implies that $$ b_{n+1}=-\frac12 b_n\tag{2} $$ [Note: One can easily find the form of $b_n$ by setting $a_{n+1}+b=-\frac12(a_n+b)$.] Now one only needs to show that $$ \lim_{n\to\infty}b_n=0. $$ But (2) gives: $$ b_{n}=q^{n-3}b_3,\quad n\geqslant 3\tag{3} $$ where $|q|=\frac12$.

---

Your $(b_n)$ and $(c_n)$ can be read from (3) if you want.

---

[Added later.] Yet, there is another way to do this problem using linear algebra. Noticing that $$ a_{n+1}=\frac{1}{2}a_n+\frac12a_{n-1},\quad a_{n+2}=\frac34a_n+\frac14a_{n-1},\quad n\ge 1, $$ one can write $b_{n+2}=Ab_n$ where $b_n=(a_{n-1},a_n)^T$ and $$ A=\frac14\begin{pmatrix} 2&2\\ 1&3 \end{pmatrix} =SJS^{-1} $$ where $$ J=\begin{pmatrix} \frac14&0\\ 0&1 \end{pmatrix},\quad S=\begin{pmatrix} -2&1\\ 1&1 \end{pmatrix},\quad S^{-1}=\frac13\begin{pmatrix} -1&1\\ 1&2 \end{pmatrix}. $$ Now, $$ b_{2n+1}=A^{n}b_1,\quad n\ge1.\tag{4} $$ But as $n\to\infty$, $$ A^{n}=SJ^{n}S^{-1}\to S \begin{pmatrix} 0&0\\0&1 \end{pmatrix}S^{-1}= \frac13\begin{pmatrix} 1&2\\1&2 \end{pmatrix}.\tag{5} $$ Combining (4) and (5) one gets $$ (a_{2n},a_{2n+1})\to (\frac23,\frac23)\quad\text{as }n\to\infty. $$ Now you can apply the theorem you have to conclude that $$ \lim_{n\to\infty}a_n=\frac23. $$

Answer 2

Rewriting the recursion you obtain

• $a_n=\frac{a_{n-1}+a_{n-2}}{2} \Leftrightarrow a_n - \frac{1}{2}a_{n-1}-\frac{1}{2}a_{n-2}= 0$

This is a linear difference equation with the characteristic polynomial $$x^2 -\frac{1}{2}x-\frac{1}{2} = 0 \Leftrightarrow \left(x + \frac{1}{2} \right)(x-1)= 0$$ So, the general solution is $$a\cdot 1^n + b \cdot \left(-\frac{1}{2} \right)^n = a+b\left(-\frac{1}{2} \right)^n \stackrel{a_1 =0, a_2 = 1}{\Longrightarrow}\frac{2}{3} + \frac{4}{3}\left(-\frac{1}{2} \right)^n \stackrel{n \to \infty}{\longrightarrow}\frac{2}{3}$$

Answer 3

Note that $0,1,5,21,85,...$ is the sequence $\frac{4^n -1}{3}$ for $n \geq 0$. Also, $*, 2, 8,32,128,\dots = 2^{2n-1}$. Note that the missing term $*$ will fit any pattern, so we need only worry about larger values. These observations show the first sequence converges to $2/3$.

The other sequence is harder to spot the pattern: $1,3,11,43,\dots = \frac{2^{2n+1}+1}{3}$. You should have no problem finding a pattern for the denominators.

Answer 4

Guide: Define $$b_n := \frac{b_{n-1}+c_{n-1}}{2} \qquad c_n := \frac{b_{n}+c_{n-1}}{2}$$ with $b_1 := a_1$ and $c_1 := a_2$ for all $n \in \Bbb{N}$. It's not hard to see that $$b_n = a_{2n-1} \qquad c_n = a_{2n}$$ by induction.

Answer 5

We have that $$2a_n=a_{n-1}+a_{n-2}$$therefore $$2(a_n-a_{n-1})=a_{n-2}-a_{n-1}$$defining $b_n= a_n-a_{n-1}$ we obtain$$2b_n=-{b_{n-1}}$$since $b_2=a_2-a_1=1$ we have $$b_n=4(-{1\over 2})^n$$which yields to $$a_n=a_1+\sum_{k=2}^{n}b_k=\sum_{k=2}^{n}b_k$$and we can write $$\lim_{n\to \infty}a_n=\sum_{k=2}^{\infty}b_k=b_2+b_3+b_4+\cdots= {1\over 1-\left(-{1\over 2}\right)}={2\over 3}$$