This is my recursive sequence:

$a_1=\frac{1}{4};\space a_{n+1}=a_n^2+\frac{1}{4}$

for $n\ge 1$

In order to check if this converges I think I have to show that

1) The sequence is monotone increasing/decreasing

2) The sequences is bounded by some value

For 1) I am going to use the ratio test.

$\frac{a_{n+2}}{a_{n+1}}>1$ $\implies$ monotone increasing

$\frac{a_{n+2}}{a_{n+1}}<1$ $\implies$ monotone decreasing

$\frac{(a_{n+1})^2+\frac{1}{4}}{a_{n+1}}=a_{n+1}+\frac{1}{4}>0$ $\implies$monotone increasing

I am really not sure about this. How would I checkt/show it is bounded by some value?

  • $\begingroup$ How did you get $a_{n+1}+\frac{1}{4} $? $\endgroup$
    – science
    Feb 8 '15 at 15:11
  • $\begingroup$ First you should find the value that it is bounded by $L$, by using the steady state property of the limit $L = a_{n} = a_{n+1}$ and solve for $L$. Then you prove inductively that $a_n < L \rightarrow a_{n+1} < L$. $\endgroup$
    – DanielV
    Feb 8 '15 at 15:23
  • $\begingroup$ Also, the initial condition is essential for proving (1). With a different initial condition (ex, $a_1 = 1/5$), then (1) wouldn't hold. $\endgroup$
    – DanielV
    Feb 8 '15 at 15:41

$a_n$ is increasing because $$ \begin{align} a_{n+1}-a_n &=a_n^2-a_n+\tfrac14\\ &=(a_n-\tfrac12)^2\\ &\ge0 \end{align} $$ If $a_n\le\frac12$, then $$ \begin{align} a_{n+1} &=a_n^2+\tfrac14\\ &\le\frac12 \end{align} $$ This implies that $a_n$ converges to some value less than or equal to $\frac12$. Let this value be $a$, then take the limit of the recursive equation $$ \begin{align} a &=\lim_{n\to\infty}a_{n+1}\\ &=\lim_{n\to\infty}a_n^2+\tfrac14\\ &=a^2+\tfrac14 \end{align} $$ which implies that $0=a^2-a+\frac14=(a-\frac12)^2$; that is, $a=\frac12$.


$a_{n+1}-a_n=\frac{(2a_n-1)^2}{4}> 0$. So this is a monotone increasing sequence. Now to see whether the sequence is bounded or not, observe that the limiting value should satisfy $a=a^2+1/4\implies a=1/2$. So, let the sequence be unbounded. Then $\exists N$ such that $a_{N-1}\le 1/2,\ a_N>1/2$. But $a_{N}>1/2\implies a_{N-1}>1/2$ which leads to a contradiction. Hence the sequence is bounded and converges to $1/2$.


It's easy to show that it's increasing by induction; $a_2=\frac1{16}+\frac14>\frac14>a_1$, and if $a_{n+1}>a_n$ for some $n\geqslant 1$, then $$a_{n+2}=a_{n+1}^2+\frac14>a_n^2+\frac14=a_{n+1} $$ (to be pedantic, you need to prove that $a_n>0$ for all $n$, but that is straightforward). Similarly, we can show that it is bounded above; clearly $a_1<\frac12$, and if $a_n<\frac12$ for some $n\geqslant 1$ then $$a_{n+1}=a_n^2 + \frac14 < \left(\frac12\right)^2 + \frac14 = \frac12. $$ I'll leave it to you to find the limit.

  • $\begingroup$ Thanks. So the ratio test I used was correct? $\endgroup$
    – qmd
    Feb 8 '15 at 14:37
  • 1
    $\begingroup$ Well, $\frac{a_{n+2}}{a_{n+1}}>1$ if and only if $a_{n+2}>a_{n+1}$. So if you can prove that directly that works too. Generally for recursively defined sequences it is easier to show it by induction though. $\endgroup$
    – Math1000
    Feb 8 '15 at 14:39

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