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We define a sequence recursively by $$a_{n+1}=\frac{1}{4}({a_n}^2+a_n+2)~~~~~~~(a_1=3)$$

By showing $a_n$ is increasing prove that $a_n$ does not converge.

Not sure how to do this one. I tried showing that $a_{n+1}-a_n>0$ but couldn't get it to work also not sure how this would be used to imply it doesn't converge. (Although I think you can just state it is unbounded so eventually it will be bigger than any limit you could imagine)

Thanks

Ryan
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    There are increasing sequences which are convergent, e.g. $\left(-\frac 1n\right)_{n\in \mathbb N}$. In this particular case, however, if the sequence is increasing and you assume it is bounded, it follows that it converges. From this you can find the candidate limits and get a contradiction. – Git Gud Mar 21 '15 at 21:41
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    Have you tried working with $f(x):=(x^2+x+2)/4$? – Alp Uzman Mar 21 '15 at 21:42
  • How is the sequence you said increasing? It goes $1, \frac{1}{2}, \frac{1}{3},...$ so it is decreasing no? – Ryan Mar 21 '15 at 21:44
  • @Ryan Typo, I fixed it. – Git Gud Mar 21 '15 at 21:45
  • you can also think about $\frac{n}{n+1}$ – benji Mar 21 '15 at 21:45

2 Answers2

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rewrite $a_{n+1}=\frac{1}{4}[a_n(a_n+1)]+\frac12$ prove by induction that $a_n+1\ge4$ and $a_{n+1}>a_n\ge3$.
(Do not forget to use that $a_1=3$ for otherwise the proof won't work.)

Once you do that, if the sequence was convergent then both $a_n$ and $a_{n+1}$ would converge to the same limit $L>3$. You would get $L=\frac{1}{4}(L^2+L+2)$. Solve this quadratic equation for $L$ and figure that the roots are too small to be a limit of numbers greater than $3$.

Alternatively, don't bother with the quadratic equation about $L$, but prove by induction that $a_n+1\ge4$, $a_{n+1}>a_n\ge3$, and $a_{n+1}\ge a_n+\frac12$. Then $a_{n+1} - a_n\ge \frac12$ for all $n$, which in particular implies that the sequence $a_n$ is not Cauchy, and cannot possibly converge. It also shows directly that $a_n\to\infty$, that is, $a_n$ diverges to infinity (which you could have also concluded, indirectly, using that $a_n$ is increasing but has no limit, after you found the possible roots for $L$).

Mirko
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$\begin{array}\\ a_{n+1} &=\frac{1}{4}({a_n}^2+a_n+2)\\ &=\frac{1}{16}(4{a_n}^2+4a_n+8)\\ &=\frac{1}{16}(4{a_n}^2+4a_n+1+7)\\ &=\frac{1}{16}((2{a_n}+1)^2+7)\\ \end{array} $

Therefore $2a_{n+1}+1 =\frac{1}{8}((2{a_n}+1)^2+7)+1 =\frac{1}{8}((2{a_n}+1)^2+15) $.

Letting $b_n =2a_n+1 $, $b_{n+1} =\frac{1}{8}(b_n^2+15) $.

Since $b_1 =2a_1+1 =7 $, $b_2 =\frac{49+15}{8} =8 $, $b_3 =\frac{64+15}{8} =\frac{79}{8} > 9 $.

If $b_n \ge 8+k $ where $k \ge 1$ (which is true for $n=3$), then $b_{n+1} =\frac{1}{8}(b_n(8+k)+15) =\frac{1}{8}(8 b_n+kb_n+15) =b_n+\frac{1}{8}(kb_n+15) =b_n+\frac{1}{8}(k(8+k)+15) $ so $b_{n+1}-b_n =\frac{1}{8}(k(8+k)+15) = k+\frac{1}{8}(k^2+15) \ge k+2 $.

Therefore $b_n$ is unbounded and strictly increasing and so is $a_n$.

marty cohen
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