# Solve this inequality with nested radicals (possibly by induction)

I tried to solve this problem by induction but didn't succeed. Given the series

$$a_n = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{... + \sqrt{n}}}}}$$

Prove that $$a_n < 2 (\forall n \in \mathbb{N^*})$$

Now I thought that maybe I could find a reccurence formula. I haven't found one. Another way I tought of was squaring both sides and substracting the number before the radical n times but that made it more complicated. Can someone lend me a hand on this?

Fix $$x \ge 0$$, and define $$(a_n)$$ by $$a_n = \sqrt{x^2 + \sqrt{x^4 + \sqrt{x^8 + \sqrt{\cdots + \sqrt{x^{2^n}} } } } }$$ Then for all $$n > 1$$, we have the relation $$a_n^2=x^2+xa_{n-1}$$ Now let $$x = \sqrt{5}-1$$.

Claim:$$\;a_n < 2$$, for all $$n$$.

Proceed by induction on $$n$$.

For $$n=1$$, we have $$a_1=\sqrt{x^2}=x = \sqrt{5}-1< 2$$.

Suppose $$a_n < 2$$, for some positive integer $$n$$.

Then we get $$a_{n+1}^2 = x^2+xa_n < x^2+2x = (\sqrt{5}-1)^2+2(\sqrt{5}-1)= 4$$ so $$a_{n+1} < 2$$, which completes the induction.

Next, compare $$x^{2^n}$$ and $$n$$ . . .

Claim:$$\;x^{2^n} > n$$, for all $$n$$.

Proceed by induction on $$n$$.

For $$n=1$$, we have $$x^{2^1} = x^2 = (\sqrt{5}-1)^2 > 1$$.

For $$n=2$$, we have $$x^{2^2} = x^4 = (\sqrt{5}-1)^4 > 2$$.

Suppose $$x^{2^n} > n$$, for some positive integer $$n \ge 2$$.

Then we get $$x^{2^{n+1}}=\left(x^{2^n}\right)^2 > n^2 \ge 2n > n+1$$ so $$x^{2^{n+1}} > n+1$$, which completes the induction.

Hence, for all $$n$$, we have $$\sqrt{1 + \sqrt{2 + \sqrt{3 + \sqrt{\cdots + \sqrt{n} } } } } < \sqrt{x^2 + \sqrt{x^4 + \sqrt{x^8 + \sqrt{\cdots + \sqrt{x^{2^n}} } } } } < 2$$

• Interesting. How did you find the value x = $\sqrt{5} - 1$? Commented Sep 27, 2018 at 17:50
• @Stefan Octavian: $\sqrt{5}-1$ is the positive root of the equation $x^2+2x=4$, which suffices to allow the first induction to work. But in fact, any value of $x$ in the interval $\sqrt[4]{2}< x \le \sqrt{5}-1$ would work just as well. Commented Sep 27, 2018 at 18:17

It is well known that the iteration $$x\mapsto \sqrt{1+x}$$ with starting point $$x=0$$ converges to $$\varphi=\frac{1+\sqrt{5}}{2}$$, i.e. $$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}=\frac{1+\sqrt{5}}{2}.\tag{1}$$ If we multiply both sides by $$2^{1/4}$$ we get

$$\sqrt{2^{1/2}+\sqrt{2+\sqrt{4+\sqrt{16+\ldots}}}}=\frac{1+\sqrt{5}}{2^{3/4}}<2\tag{2}$$ and the LHS of $$(2)$$ is blatantly larger than any $$a_n$$ since $$2^{2^{n-2}}\geq n$$ for any $$n\geq 1$$.
Actually $$\lim_{n\to +\infty} a_n = \sup_{n\geq 1}a_n \approx 1.75793\ldots < \frac{2495-\sqrt{5}}{1418}.\tag{3}$$

$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{... + \sqrt{n}}}}}\le \sqrt{1\sqrt{2\sqrt{3\sqrt{...\sqrt{n}}}}} = \prod_{k=1}^nk^{\frac{1}{2^k}} < 1.7$$