Proving that the sequence $a_0=0,\ a_{n+1}=(a_n)^2+\frac{1}{4}$ converges to $\frac{1}{2}$

Question

I am reading a real analysis book which in the chapter about sequences and series (which does not include anything on recurrence relations) asks to prove that the sequence $a_0=0,\ a_{n+1}=(a_n)^2+\frac{1}{4}$ is increasing, bounded and then asks to compute its limit.

I have managed to prove that $\{a_n\}$ is increasing and bounded above as follows. Clearly $a_n\geq 0$ for all $n\in\mathbb{N}$ and by AM-GM inequality we get that $a_{n+1}=(a_n)^2+\frac{1}{4}\geq 2\sqrt{(a_n)^2\cdot\frac{1}{4}}=a_n$ for all $n\in\mathbb{N}.$ Also, $a_0=0<\frac{1}{2}$ and if $a_n<\frac{1}{2}$ then $a_{n+1}=(a_n)^2+\frac{1}{4}<\left(\frac{1}{2}\right)^2+\frac{1}{4}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ so by induction it follows that $a_{n}<\frac{1}{2}$ for all $n\in\mathbb{N}$. So, since $\{a_n\}$ is monotonically increasing and bounded above, it must converge to some $L\in\mathbb{R},\ L\leq\frac{1}{2}.$

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Now, by numerical inspection it seems that $a_n\xrightarrow[]{n\to\infty}\frac{1}{2}$ but I haven't been able to prove this last statement; I know that I have to prove that for every $\varepsilon>$ there exists $N\in\mathbb{N}$ such that $a_N>\frac{1}{2}-\varepsilon$ but right now I can't see how to do this. The best I have been able to obtain so far is that since $a_n\geq\sum\limits_{j=1}^{n}\left(\frac{1}{4}\right)^j$ it must be $\lim\limits_{n\to\infty}a_n\geq\sum\limits_{j=1}^{\infty}\left(\frac{1}{4}\right)^j=\frac{1}{3}.$

So, I would be interested in knowing how to prove this last statement. Thanks.

Answer 1

The hard part, which you've done, is showing that the limit exists.

Then, the following proposition yields the equation $f(L) = L$. In general, such an equation might be difficult to solve, but in your case, it's straightforward.

Claim. Suppose a sequence is defined by a recurrence $a_{n+1} = f(a_n)$, for some continuous $f$. If we know that $a_n \to L$ as $n \to \infty$, then $f(L) = L$.

Hint: The crucial observation is the fact that the limits of a sequence and of a shifted version of that sequence are the same: For any $k \in \mathbb{N}$, $$ \lim_{n \to \infty} a_{n+k} = \lim_{n \to \infty} a_n. $$

Try to prove the claim, and click to reveal the proof if you get stuck.

\begin{align} f(L) &= f\Bigl(\,\lim_{n \to \infty} a_n \Bigr) \\ &= \lim_{n \to \infty} f(a_n) \\ &= \lim_{n \to \infty} a_{n+1} \\ &= \lim_{n \to \infty} a_n \\ &= L. \end{align}

Answer 2

It looks like you’ve already shown that the limit L exists, but you’re struggling with how to find the exact value of the limit.

I assume you can use the following conclusion: If a sequence converges, then it also converges in the Cauchy sense.

it means: $\forall\epsilon>0,\exists N\in Z^+,s.t. \forall n>N,|a_{n+1}-a_n|<\epsilon$

so For a monotonically increasing sequence, we have the following:

$a_n+\epsilon>a_n^2+\frac 14$

Now, all we need to do is utilize the premise that epsilon can be made arbitrarily small to solve the above equation

then we have $|a_n-\frac 12| <\sigma$, and $\sigma$ is a small num determined by $\epsilon$