# What is the value of S=sqrt(6+2sqrt(7+3sqrt(8+4sqrt(9+cdots))))?

Jun 19, 2017

$4$

#### Explanation:

First, we can arrange this relationship in the form

$f \left(x\right) = \sqrt{x + 4 + x \sqrt{x + 5 + \left(x + 1\right) \sqrt{x + 6 + \left(x + 2\right) \sqrt{\ldots}}}}$

Now we will present a simple guess to find the form of $f \left(x\right)$

For more details I recommend the lecture of the marvelous work of Ramanujan. Himself submitted the present exercise in the Journal of the Indian Mathematical Society and the solution was subsequently furnished also. The general theory of this type of recurrence formulas appears in Ramanujan's Notebooks - Part II page 108 in a book published by Springer Verlag and edited by Bruce C. Berndt.

We know that

$f \left(x\right) < f \left(x + 1\right) < f \left(x + 2\right) < \cdots < f \left(x + n\right)$ and also that

$f \left(0\right) = \sqrt{4} = 2$

so at this point we propose

$f \left(x\right) = {c}_{0} + {c}_{1} x$

Substituting into the general recurrence equation as

$f \left(x\right) = \sqrt{x + 4 + x f \left(x + 1\right)}$ or

${\left({c}_{0} + {c}_{1} x\right)}^{2} = x + 4 + x \left({c}_{0} + {c}_{1} \left(x + 1\right)\right)$

and now grouping coefficients,

${x}^{2} \left({c}_{1}^{2} - {c}_{0}^{2}\right) + \left(2 {c}_{0} {c}_{1} - 1 - {c}_{0} - {c}_{1}\right) x + \left({c}_{0}^{2} - 4\right) = 0$

Solving now

$\left\{\begin{matrix}{c}_{1}^{2} - {c}_{0}^{2} = 0 \\ 2 {c}_{0} {c}_{1} - 1 - {c}_{0} - {c}_{1} = 0 \\ {c}_{0}^{2} - 4 = 0\end{matrix}\right.$

we get at ${c}_{0} = 2$ and ${c}_{1} = 1$

$f \left(x\right) = 2 + x$

so finally

$f \left(2\right) = 4$