1
$\begingroup$

Consider sequence $a_n$ defined by
$$a_1=2 $$
$$7a_{n+1} = a_n^2+3, n\ge2$$

  • Show that $\frac{1}{2}\lt a_n \lt 3$
  • Prove that it is increasing.
  • Find its limit as $n\to+\infty$.

I know how to prove the first part by induction. However, I don't know how to prove the second part.
And if the sequence is both bounded and increasing, which is monotonic, it is going to converge, right? But I don't think it is going to converge at all.

$\endgroup$
  • 3
    $\begingroup$ a_2=1, how is it increasing? $\endgroup$ – Joel Moreira Nov 10 '15 at 3:10
  • $\begingroup$ It is not an increasing sequence $\endgroup$ – Kushal Bhuyan Nov 10 '15 at 3:11
  • 2
    $\begingroup$ Wait, the formula is only for $n\geq2$. But then what is $a_2$? $\endgroup$ – Joel Moreira Nov 10 '15 at 3:11
  • $\begingroup$ I am not sure. That's what it said in the question. I have to ask my professor about it. $\endgroup$ – abuchay Nov 10 '15 at 3:12
  • $\begingroup$ the formula should start with $n\geq 1$ or there is a value for $a_2$ $\endgroup$ – Kushal Bhuyan Nov 10 '15 at 3:15
1
$\begingroup$

First, for any value of $a_n$, $a_{n+1}>3/7$. So, the sequence is bounded below by $3/7$. Second, if $a_n<3$, then $a_{n+1}<12/7<3$.

Therefore, if the initial value $a_1<3$, then $3/7<a_n<12/7<3$ for $n\ge2$.

Next, let's examine the first difference $a_{n+1}-a_n$. We have

$$\begin{align} a_{n+1}-a_n&=\frac{a_n^2+3}{7}-a_n\\\\ &=\frac{a_n^2+7a_n+3}{7}\\\\ &=\frac17 \left(a_n-\frac{7+\sqrt{37}}{2}\right)\left(a_n-\frac{7-\sqrt{37}}{2}\right)\\\\ \end{align}$$

Therefore, $a_{n+1} < a_n$ whenever $0<\frac{7-\sqrt{37}}{2}<a_n<\frac{7+\sqrt{37}}{2}<7$.

Inasmuch as $a_1=2$ $a_n$ is a monotonically decreasing sequence that is bounded below by $3/7$. Then, by the Monotone Convergence Theorem, $a_n$ converges.

Finally, letting $\lim_{n\to \infty}a_n=L$, we can write

$$\begin{align} \lim_{n\to \infty}a_{n+1}&=\lim_{n\to \infty}\frac{a_n^2+3}{7}\\\\ L&=\frac{L^2+3}{7}\\\\ L^2-7L+3&=0\\\\ L&=\bbox[5px,border:2px solid #C0A000]{\frac{7-\sqrt{37}}{2}} \end{align}$$

where we ruled out the solution $L=\frac{7+\sqrt{37}}{2}>13/2$.

$\endgroup$
0
$\begingroup$

It is converging and the limit is going to be the smaller solution of $x^2-7x+3=0$, which is ${7-\sqrt{37}\over2}$.

Also it is strictly decreasing. Proof:

Base case when $n=1,2$ is trivial and $a_n>a_{n+1}\implies {{a_n}^2+3\over7}>{{a_{n+1}}^2+3\over7}\implies a_{n+1}>a_{n+2}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.