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Question: Let $a_1=0, a_2=3$, and, for all $n \geq 3$ let $a_n = \frac{1}{2} (a_{n-1}+a_{n-2})$

The sequence $(a_n)$ is said to be defined recursively.

on $n$, show that, for all $n \geq 2$,

$$a_n=2+4 \left(\frac{-1}{2}\right)^n$$

And deduce that $(a_n) \rightarrow 2$

My attempt:

$ a_1 = 0 , a_2 = 3 , \ldots $

$\forall n \geq 3$ let $a_n = \frac{1}{2}(a_{n-1} + a_{n-2})$

Show inductively that $\forall n \geq 2$

$ a_n = 2 + 4(- \frac{1}{2})^n $

And deduce that $(a_n) \rightarrow 2$

So $\ldots$

let $n$ = $3$ then $a_3 = \frac{1}{2}(a_2 + a_{1}) = \frac{1}{2}(3 + 0) = \frac{3}{2}$

This is equal to $2 + 4(- \frac{1}{2})^3$, so this holds.

Assume that this holds up to $n = k$, then for $n = k + 1$;

$a_{k + 1} = \frac{1}{2}(a_k + a_{k-1})$

$a_k$ is known as $a_k = 2 + 4(- \frac{1}{2})^k$

And $a_{k - 1}$ is also known as $a_{k - 1}= 2 + 4(- \frac{1}{2})^{k - 1}$

so I then have

$2a_{k+1} = (a_k + a_{k-1})$, and subbing in the values above for $a_k$ and $a_{k-1}$ gives

$2a_{k+1} = ((2 + 4(- \frac{1}{2})^k) + (2 + 4(- \frac{1}{2})^{k - 1}))$

Which can be simplified to

$a_{k+1} = (1 + 2(- \frac{1}{2})^k + 1 + 2(- \frac{1}{2})^{k - 1})$

and

$a_{k+1} = 2 + 2(- \frac{1}{2})^k + 2(- \frac{1}{2})^{k - 1}$

Using exponent laws for $ 2(- \frac{1}{2})^{k - 1}$ gives

\begin{equation*} \begin{aligned} 2(- \frac{1}{2})^{k - 1} &= 2(-\frac{1}{2})^{k} \times (- \frac{1}{2})^{-1} \\ &= 2(-\frac{1}{2})^{k} \times (-2) \\ &= -4(-\frac{1}{2})^{k} \\ \end{aligned} \end{equation*}

Subbing this back into $a_{k+1} = 2 + 2(- \frac{1}{2})^k + 2(- \frac{1}{2})^{k - 1}$ as

\begin{equation*} \begin{aligned} a_{k+1} &= 2 + 2(- \frac{1}{2})^k -4(-\frac{1}{2})^{k} \\ &= 2 - 2(- \frac{1}{2})^k \end{aligned} \end{equation*}

If the induction holds then

$ a_{k + 1} = 2 - 2(- \frac{1}{2})^k = 2 + 4(- \frac{1}{2})^{k + 1}$

Using power laws on $ 4(- \frac{1}{2})^{k + 1}$ gives

\begin{equation*} \begin{aligned} 4(- \frac{1}{2})^{k + 1} &= 4(- \frac{1}{2})^{k } \times (- \frac{1}{2})^{1} \\ &= 4(- \frac{1}{2})^{k } \times (- \frac{1}{2}) \\ &= -2(- \frac{1}{2})^{k } \\ \end{aligned} \end{equation*}

Using this result gives

\begin{equation*} \begin{aligned} a_{k + 1} &= 2 - 2(- \frac{1}{2})^k \\ &= 2 + 4(- \frac{1}{2})^{k + 1} \\ &= 2 -2(- \frac{1}{2})^{k } \\ \end{aligned} \end{equation*}

As required.

Therefore by the inductive hypothesis $\forall n \geq 2, a_n = 2 + 4(- \frac{1}{2})^n $

Deduce that $(a_n) \rightarrow 2$

If $(a_n) \rightarrow 2$ then as $n \rightarrow \infty, a_n \rightarrow 2$. So the limit of the sequence is $2$.

Using the definition of convergence:

$\forall \epsilon > 0$ there exist natural numbers $N,n$ where $n>N$ such that $|2 + 4(- \frac{1}{2})^n - 2| < \epsilon$

Which simplifies to $|4(- \frac{1}{2})^n| < \epsilon$ and $4(\frac{1}{2})^n < \epsilon$

Then, given $\epsilon > 0$ choose $N$ such that the minimum value of $N$ is $\log_2(\frac{4}{\epsilon})$ such that for all $n > N, a_n < \epsilon$

This can be seen as

\begin{equation*} \begin{aligned} N &> \log_2(\frac{4}{\epsilon}) \\ 2^N &> \frac{4}{\epsilon} \\ \frac{1}{2^N} &< \frac{\epsilon}{4} \\ \frac{4}{2^N} &< \epsilon \\ \end{aligned} \end{equation*}

As required.

This proves that $a_n \rightarrow 2$

Not sure if this is correct but could someone please check and if there is an easier way of proving this let me know!

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  • $\begingroup$ Looks correct... $\endgroup$ – Simply Beautiful Art Jan 5 '17 at 22:56
  • $\begingroup$ For the limit, can't you use $\left( -1 \over 2 \right)^n \longrightarrow 0$ ? $\endgroup$ – Blencer Jan 5 '17 at 22:57
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Hint:  rewrite the recursion as $2(a_n - a_{n-1})=-(a_{n-1}-a_{n-2})$. With $b_n = a_n - a_{n-1}\,$: $$b_n = -\,\frac{1}{2} \,b_{n-1}$$

Therefore $b_n$ is a geometric progression, and $a_n=b_n+a_{n-1}=b_n+b_{n-1}+a_{n-2}=\cdots$

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By a well known theorem the general solution of $a_n = \dfrac{1}{2}(a_{n-1} + a_{n-2})$ can be found by means of the characteristic equation: $$\lambda^2=\dfrac{1}{2}(\lambda+1)=0\Leftrightarrow \ldots \Leftrightarrow \lambda=1\vee\lambda=-1/2,\text{ so }a_n=C_1(1)^n+C_2\left(-\dfrac{1}{2}\right)^n.$$ $$n=1\Rightarrow 0=C_1-\dfrac{1}{2}C_2,\quad n=2\Rightarrow 3=C_1+\dfrac{1}{4}C_2.$$ Solving the system we get $C_1=2,\;C_2=4$, so $$\boxed{\;a_n=2+4\left(-\dfrac{1}{2}\right)^n\;}$$

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  • $\begingroup$ Excellent answer!! :) $\endgroup$ – manooooh Sep 16 '18 at 4:00
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    $\begingroup$ Thanks manooooh. See you in rinconmatematico. :) $\endgroup$ – Fernando Revilla Sep 20 '18 at 7:56

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