Nested Horror

This is a known problem given by Ramanujan. We will investigate a more general case:

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

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

$f(x)^2 = ax + (n+a)^2 + xf(x+n)$

Letting $f(x) = x+n+a$ , we find that this satisfies our question. (In particular, $(x+n+a)^2 = ax + (a+n)^2 + x(x+2n+a)$ )

Thus, our question gives $a = 0, n = 1, x = 2$ , so $f(x) = 0+1+2 = 3$ .

First, if we write $n \times (n+2)$ = $n \times \sqrt{1+(n+1)(n+3)}$ ),

And we defined F(n)= $n \times (n+2)$ .

So we have that,

F(n)= $n \times \sqrt{1+(n+1)(n+3)}$

F(n)= $n\times \sqrt { 1+f\left( n+1 \right) }$

F(n)= $n\times \sqrt { 1+(n+1)\times \sqrt { 1+(n+2)\times (n+4) } }$

F(n)= $n\times \sqrt { 1+(n+1)\times \sqrt { 1+(n+2)\times \sqrt { 1+(n+3)\times (n+5) } } }$

So, $n\times \sqrt { 1+(n+1)\times \sqrt { 1+(n+2)\times \sqrt { 1+... } } }$ = $n\times (n+2)$

Then, if we replace $n=1$

$1\times (1+2)$ = $1\times \sqrt { 1+2\times \sqrt { 1+3\times \sqrt { 1+... } } } )$

$3=1\times \sqrt { 1+2\times \sqrt { 1+3\times \sqrt { 1+... } } } )$

Ramanujan's solution

Familiar Identity Ramanujan Nested Radical

The problem above can simplify to (1 + 2x4)^(1/2) = 9^(1/2) = 3. ANSWER : 3

Well took a calculator ; first sqrt(1+4)=sqrt(5)*3= 6.708 ; Then added 1 == 7.708 again sqrt(7.708)=2.7763; multiplied it by 2 to get 5.552 ; added 1 6.552 then sqrt(6.552) =2.559... Since the sum is a positive sum and root values cannot be accurately typed with a computer(not perfect squares), I guessed the answer to be 3.... :)

Try to use the calculator and determine the values by increasing the numbers as shown in the sequence and you will notice that the values are approaching 3 but doesn't equal it!