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

<br /> x_{n} = \sqrt{1 + n \sqrt{1 + (n +1) \sqrt{1 + \ldots}}}<br />

Then, we have the backwards recursive formula:

<br /> x_{n} = \sqrt{1 + n x_{n + 1}}<br />

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:

<br /> x_{n} \sim -\frac{1}{n}, \; n \rightarrow \infty<br />

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

<br /> x_{n} \sim A \, n^{\alpha}, \; n \rightarrow \infty, \; \alpha &gt; 0<br />

Then, we have:

<br /> A n^{\alpha} \sim \sqrt{1 + n A (n + 1)^{\alpha}} \sim \sqrt{A} \, n^{\frac{1 + \alpha}{2}}<br />

which means:

<br /> \alpha = \frac{1 + \alpha}{2} \Rightarrow \alpha = 1<br />

and

<br /> A = \sqrt{A} \Rightarrow A = 0 \vee A = 1<br />

So, we may conclude:

<br /> x_{n} = n (1 + y_{n}), \; y_{n} \rightarrow 0, \; n \rightarrow \infty<br />

The backwards recursive relation for y_{n} is:

<br /> n (1 + y_{n}) = \sqrt{1 + n \cdot n (1 + y_{n + 1})}<br />

<br /> y_{n} = \sqrt{1 + y_{n + 1} + \frac{1}{n}} - 1<br />

<br /> y_{n + 1} = y_{n} (2 + y_{n}) - \frac{1}{n}<br />

Define a discrete z-transform:

<br /> Y(z) \equiv \sum_{n = 1}^{\infty}{y_{n} z^{-n}}<br />

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

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

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