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

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 $$L\in\mathbb{R},\ L\leq\frac{1}{2}.$$

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.

• You have shown that it is increasing and bounded, hence it approaches a limit $L$. But then we have $L=L^2+\frac 14\implies L=\frac 12$.
– lulu
Commented Nov 7, 2023 at 23:43
• @lulu : ah, I can't believe I didn't see this. Thank you very much; if you would write your comment as an answer, I would gladly accept it. Commented Nov 7, 2023 at 23:44
• You can post your own solution, you did $99\%$ of the work on it, after all.
– lulu
Commented Nov 7, 2023 at 23:47

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}

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 $$\epsilon$$