# Find the sum $\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}}$

Okay so this can be written as $$\sqrt{5+\sqrt{(5+6)+\sqrt{(5+6+8)+\sqrt{(5+6+8+10)+\sqrt{(5+6+8+10+12)\cdots}}}}}$$ Putting it as $$y$$ and squaring both sides doesn't seem to help, and I don't know what else can be done.

• Numerically, it seems to converge to 3. – greelious Jan 20 at 3:16
• Thanks, but I actually need to compute it theoretically. – Goal123 Jan 20 at 3:17
• Please give the rule for the coefficients $5, 11, 19, \ldots$ in your formula (which, by the way, is not a sum). – Rob Arthan Jan 20 at 3:31
• The terms seem to be given by $n^2 + 3n + 1$ for $n \geq 1$. – rwbogl Jan 20 at 3:40
• Just a nitpick on the title: this is not actually a sum. – Cheerful Parsnip Jan 20 at 5:49

## 3 Answers

We may adopt the technique for Ramanujan's infinite radical. Let $$p(x) = x^2 + 3x + 1$$ and define $$F : [0, \infty) \to [0, \infty)$$ by

$$F(x) = \sqrt{p(x) + \sqrt{p(x+1) + \sqrt{p(x+2) + \cdots }}}$$

Then $$F$$ solves

$$F(x)^2 = p(x) + F(x+1).$$

Now we make an ansatz that $$F(x)$$ takes the form $$F(x) = ax + b$$. Plugging this and comparing coefficients shows that

$$F(x) = x + 2$$

solves the functional equation. Finally, since $$(p(1), p(2), p(3), \cdots) = (5, 11, 19, \cdots)$$, the infinite radical in question corresponds to the case $$x = 1$$, giving

$$\sqrt{5 + \sqrt{11 + \sqrt{19 + \cdots}}} = F(1) = 3.$$

Rigorous justification. Let $$\mathcal{C}$$ be the set of all continuous functions $$f : [0, \infty) \to \mathbb{R}$$ such that

$$\| f\| := \sup_{x\to\infty} \left( 2^{-x/2} |f(x)| \right)$$

is finite. Notice that $$\mathcal{C}$$ is a complete normed space with respect to $$\|\cdot\|$$. Write $$p(x) = x^2 + 3x + 1$$ and define

$$\mathcal{A} = \{ f \in \mathcal{C} : f(x) \geq 0 \text{ for all } x \geq 0 \}.$$

This is a closed subset of $$\mathcal{C}$$. Now define $$\Phi : \mathcal{A} \to \mathcal{A}$$ by

$$\Phi[f](x) = \sqrt{p(x) + f(x+1)}.$$

If $$f \in \mathcal{A}$$, then $$2^{-x/2}|\Phi[f](x)| \leq 2^{-x/2}\sqrt{p(x) + 2^{(x+1)/2}\|f\|}$$ shows that $$\|\Phi[f]\| < \infty$$, hence $$\Phi$$ is well-defined. Moreover, if $$f, g \in \mathcal{A}$$, then

\begin{align*} 2^{-x/2} \left| \Phi[f](x) - \Phi[g](x) \right| &= 2^{-x/2} \cdot \frac{\left| f(x+1) - g(x+1) \right|}{\sqrt{p(x) + f(x+1)} + \sqrt{p(x) + g(x+1)}} \\ &\leq 2^{-x/2} \cdot \frac{2^{(x+1)/2} \| f - g \|}{2} \\ &= \frac{1}{\sqrt{2}} \| f - g \|. \end{align*}

So $$\Phi$$ is a contraction mapping on $$\mathcal{A}$$, and hence, by the contraction mapping theorem,

• There exists a unique $$F \in \mathcal{A}$$ for which $$\Phi[F] = F$$, and
• Such $$F$$ is realized as the limit $$\Phi^{\circ n}[f]$$ as $$n\to\infty$$ for arbitrary initial choice $$f \in \mathcal{A}$$.

Finally, we already know that $$F(x) = x+2$$ is an element of $$\mathcal{A}$$ that solves $$\Phi[F] = F$$, and therefore, $$\forall f \in \mathcal{A} \ : \quad \lim_{n\to\infty} \Phi^{\circ n}[f](x) = x+2$$

Maybe works, $$3=\sqrt{3^{2}}=\sqrt{5+4}=\sqrt{5+\sqrt{16}}=\sqrt{5+\sqrt{11+5}}$$ $$=\sqrt{5+\sqrt{11+\sqrt{25}}}=\sqrt{5+\sqrt{11+\sqrt{19+6}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{36}}}}$$ $$=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+7}}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{49}}}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+8}}}}}=\ldots$$

This is just a beautiful way of writing 3.

• Last line should have $\sqrt{41+8}$ not $\sqrt {41+7}.$ – Mohammad Zuhair Khan Jan 20 at 4:32

This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.

Set $$a_n = n^2 + 5n + 5$$ for $$n \geq 0$$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n = \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_n + (4 + n)}}}.$$

As Pablo_Lee notes, $$r_n = 3$$ for all $$n$$. To see this, observe that $$a_n + (n + 4) = (n + 3)^2$$. This allows us to "unroll" the radical back to $$a_0$$. For example, $$a_{n - 1} + \sqrt{a_n + (n + 4)} = a_{n - 1} + n + 3 = a_{n - 1} + ((n - 1) + 4) = ((n - 1) + 3)^2.$$ Therefore, \begin{align*} r_n &= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 1} + \sqrt{a_n + (n + 4)}}}} \\ &= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 1} + ((n - 1) + 4)}}} \\ &= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 2} + ((n - 2) + 4)}}} \\ &\vdots \\ &= \sqrt{a_0 + \sqrt{a_1 + 5}} \\ &= \sqrt{a_0 + 4} \\ &= \sqrt{(0 + 3)^2} \\ &= 3. \end{align*} (There is likely a snappy way to do this by induction, but I don't see it yet.)

If we are willing to define the full radical as $$\lim_{n \to \infty} r_n$$, then this should also be an acceptable answer.

Edit: For any integer $$r \geq 2$$, setting $$p_n = n^2 + (2r - 1)n + r^2 - r - 1$$ and $$q_n = n + r + 1$$ should yield, through the same arguments, $$r = \sqrt{p_0 + \sqrt{p_1 + \cdots + \sqrt{p_n + q_n}}}$$ for all $$n \geq 0$$. Note that $$p_n$$ is merely a shifted form of the Fibonacci polynomial $$n^2 - n - 1$$ at integer values.

For example, $$4 = \sqrt{11 + \sqrt{19 + \sqrt{29 + \sqrt{41 + 8}}}}.$$