Sequence

[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.