Solve Hard Sequence Problem: Find √1+√1+2√1+3√1+...

[aznluster]

Find $$\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+...}}}}$$

 

[QuantumJG]

It comes out to be 3. Why? I'm trying to find out.

 

[QuantumJG]

Ok so this is basically one of Ramanajan's identities:

$$x + n + a = \sqrt{ax + (n+a)^{2} + x \sqrt{a(x +n) + (n+a)^{2} + (x + n) \sqrt{...}}}}$$

 

[aznluster]

Wouldn't that add up to 2? Or maybe I'm doing something wrong. Is there a proof of that identity somewhere?

 

[Dickfore]

Let

$$x_{n} = \sqrt{1 + n \sqrt{1 + (n +1) \sqrt{1 + \ldots}}}$$

Then, we have the backwards recursive formula:

$$x_{n} = \sqrt{1 + n x_{n + 1}}$$

We want to find $x_{1}$. Let us see what is the limit $x \equiv \lim_{n \rightarrow \infty}{x_{n}}$?



If we want a finite limit, we must have:

$$x_{n} \sim -\frac{1}{n}, \; n \rightarrow \infty$$

but, this is impossible since all $x_{n} > 0$. Therefore, the limit is infinite. Let us assume that is asymptotically a power law:

$$x_{n} \sim A \, n^{\alpha}, \; n \rightarrow \infty, \; \alpha > 0$$

Then, we have:

$$A n^{\alpha} \sim \sqrt{1 + n A (n + 1)^{\alpha}} \sim \sqrt{A} \, n^{\frac{1 + \alpha}{2}}$$

which means:

$$\alpha = \frac{1 + \alpha}{2} \Rightarrow \alpha = 1$$

and

$$A = \sqrt{A} \Rightarrow A = 0 \vee A = 1$$

So, we may conclude:

$$x_{n} = n (1 + y_{n}), \; y_{n} \rightarrow 0, \; n \rightarrow \infty$$

The backwards recursive relation for $y_{n}$ is:

$$n (1 + y_{n}) = \sqrt{1 + n \cdot n (1 + y_{n + 1})}$$

$$y_{n} = \sqrt{1 + y_{n + 1} + \frac{1}{n}} - 1$$

$$y_{n + 1} = y_{n} (2 + y_{n}) - \frac{1}{n}$$

Define a discrete z-transform:

$$Y(z) \equiv \sum_{n = 1}^{\infty}{y_{n} z^{-n}}$$

As we can see, $y_{1}$ is the coefficient in the Laurent series of $Y(z)$ around $z = 0$. According to the residue theorem:

$$y_{1} = \mathrm{Res}(Y(z), z = 0) = \frac{1}{2 \pi i} \, \oint_{C_{z = 0}}{Y(z) \, dz}$$

We need to find a functional equation for $Y(z)$ from the above (non-linear) recursive relation.