# Bounds for $\underbrace{\sqrt {a+\sqrt{a+\sqrt {a+…+\sqrt{a}}}}}_\text{$a$roots}$ in terms of $a$?

Inspired by questions like this and this, I have thought about what happens in the general case. My question goes like this:

What can be a suitable upper bound and a suitable lower bound to $$\underbrace{\sqrt {a+\sqrt{a+\sqrt {a+...+\sqrt{a}}}}}_\text{a roots}$$ in terms of $$a$$?

Following from the work done on the mentioned questions, an upper bound can be calculated by assuming there are infinite roots.

Upper bound:

Let $$x={\sqrt {a+\sqrt{a+\sqrt {a+\sqrt {a+\ldots}}}}}$$

$$\therefore x^2=a+{\sqrt {a+\sqrt{a+\sqrt {a+\sqrt{a+\ldots}}}}}$$ $$x^2=a+x$$ $$x^2-x-a=0$$ $$\therefore x=\frac {1 \pm \sqrt {4a+1}}2$$

$$a$$ has to be positive for $$x \in \mathbb R$$ and $$x \ge 0$$ to be a genuine answer. So $$x=\frac {1 + \sqrt {4a+1}}2$$. (I am not exactly sure whether this assumption is valid, please help me out if it isn't).

Lower bound:

Predictably $$\sqrt a$$ is a lower bound but as one of the answers to the mentioned, it is a bad lower bound.

The answer uses $$\sqrt {a +\sqrt a}$$ as a more suitable lower bound. It may not always work but is the best combination of an accurate and easy to calculate lower bound. Obviously $$\underbrace{\sqrt {a+\sqrt{a+\sqrt {a+...+\sqrt{a}}}}}_\text{a-1 roots}$$ is a much better lower bound except that it is almost worth it to calculate the one extra term than to calculate this very accurate lower bound.

So, $$\sqrt {a +\sqrt a} \le \underbrace{\sqrt {a+\sqrt{a+\sqrt {a+...+\sqrt{a}}}}}_\text{a roots} \le \frac {1 + \sqrt {4a+1}}2$$ is a good enough technique to find the upper bounds and lower bounds to $$\underbrace{\sqrt {a+\sqrt{a+\sqrt {a+...+\sqrt{a}}}}}_\text{a roots} \$$ (if my calculation is right) so no one has to bother posting the same question for the upcoming years.

Is this technique correct? Is there any better technique that produces better bounds? I have assumed nothing about when $$a$$ is large and when $$a$$ is small. Does that make a difference? Thanks in advance!

Similar to my post an hour ago. Yes, you are correct.

$$\sqrt{a+\sqrt a}<=\sqrt{a+\sqrt{a+\sqrt{...+a}}}<=\sqrt{a+\sqrt{a+\sqrt{a+...}}}$$ Here is the correct way to bound.

• Is there any better technique? When will $\sqrt {a+\sqrt {a+\sqrt a}}$ be a better (yet simpler) lower bound than $\sqrt{a+\sqrt a}$? – Mohammad Zuhair Khan Jan 21 '18 at 5:56
• Honestly, the number of a in the lower bound just has to be less than the number of a in the thing you are trying to bound. In this case I am merely making a general form – QuIcKmAtHs Jan 21 '18 at 5:57
• If $a$ is large (or small) is there any difference? – Mohammad Zuhair Khan Jan 21 '18 at 6:00

By induction for $a>0$ we obtain: $$\sqrt{a}<\frac{1+\sqrt{1+4a}}{2}$$ and $$\sqrt{a+\sqrt{a+..}}<\sqrt{a+\frac{1+\sqrt{1+4a}}{2}}=\sqrt{\frac{1+4a+2\sqrt{1+4a}+1}{4}}=\frac{1+\sqrt{1+4a}}{2}$$

• Is there any better technique? When will $\sqrt {a+\sqrt {a+\sqrt a}}$ be a better (yet simpler) lower bound than $\sqrt{a+\sqrt a}$? – Mohammad Zuhair Khan Jan 21 '18 at 5:56
• For the upper bound it's the best. – Michael Rozenberg Jan 21 '18 at 5:58
• Thanks for the help! I almost feel sorry there will be no 'Welcome $2019$' question. – Mohammad Zuhair Khan Jan 21 '18 at 6:05
Assuming $a\geq 1$, let $f(x)=\sqrt{a+x}$, $a_0=0$ and $a_{n+1}=f(a_n)$.
Since $f(x)$ is a contraction on $\mathbb{R}^+$ (we have $\left|\,f'(x)\right|<1$), by the Banach fixed point theorem the sequence $\{a_n\}_{n\geq 0}$ converges to the only fixed point of $f$, namely $L=\frac{1+\sqrt{4a+1}}{2}$. Additionally $\{a_n\}_{n\geq 0}$ is trivially increasing, hence $L-a_n$ is positive and can be bounded by exploiting the Taylor series of $f(x)$ centered at $x=L$. We have linear convergence, meaning that $$L-a_n\approx C\,f'(L)^n$$ for large values of $n$.