My favorite nested radical identity

We can write this nested radical as $g(x) = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+... }}}}$ Squaring both sides gives us $[g(x)]^2 = 1 + xg(x+1)$ We can tease out a few hints as to what $g(x)$ is. For example, we can see that $g(0)=1$ . We can also see that squaring $g(x)$ is on the same order as multiplying by $x$ , which tells us that if $g(x)$ is a polynomial, then it must be a first-order polynomial. Thus, we can guess that it has a solution of the form $g(x) = ax+1$ . Let's do some quick work to guarantee that this is right. $(ax+1)^2 = 1 + x(ax+a+1)$ $a^{2}x^{2} + 2ax + 1 = 1 + ax^2 + ax + x$ Which is only solved when $a=1$ . Thus, $g(x) = x+1$ solves this equation. Finally $g(4) = \sqrt{1+4\sqrt{1+5\sqrt{1+...}}} = 1+4 = 5$

Ramanujan discovered in 1911 that:

$x+m+a = \sqrt{ax+(m+a)^2+x\sqrt{a(x+m)+(m+a)^2+(x+m)\sqrt{a(x+2m)+(m+a)^2+(x+2m)\sqrt \cdots}}}$

Putting $x=4$ , $m=1$ and $a=0$ , we have:

$4+1+0 = \sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt {1+\cdots}}}}$

Therefore, $n = \boxed 5$ .

Proof: Consider $f(x) = x + 1$ , then $f(x) = \sqrt{(x+1)^2} = \sqrt{1+x(x+2)} = \sqrt{1+xf(x+1)}$ . $\implies f(x+1) = \sqrt{1+(x+1)f(x+2)}$ , $f(x+2) = \sqrt{1+(x+2)f(x+3)}$ ... Therefore, $x + 1 = \sqrt{1+x\sqrt{1 + (x+1)\sqrt{1+(x+2)\sqrt {1+\cdots}}}}$ . Putting $x=4$ , we have $\boxed 5 = \sqrt{1+4\sqrt{1 + 5\sqrt{1+6 \sqrt {1+\cdots}}}}$ .

recall that
$a^2-b^2=(a-b)(a+b)$ define
$f(x) = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+...}}}$ so that $n=f(4)$ now
$f(x) = \sqrt{1+xf(x+1)}$ suppose $f(x) = x+1$ , this satisfies the previous recurrence relation, as
$RHS = \sqrt{1+x(x+2)} = \sqrt{1+(x+1-1)(x+1+1)} = \sqrt{1+(x+1)^2-1} = \sqrt{(x+1)^2} = x+1 = LHS = x+1$ , thus $f(4) = 4 + 1 = 5$ is a possible value for $n$

Solve for $x$ in this equation:

$3 = \sqrt{1+2\sqrt{1+3x}}$

and get $x=5$

This is based on one of Ramanujan's most well known infinite nested radicals, which is

$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}$

If $a$ and $b$ are nonnegative numbers, then $a+b = \sqrt{b^2 + a(a + b + b)}$ Applying the above formula to $a+b + b$ gives $\sqrt{b^2 + a \sqrt{b^2 +(a+b)(a+b+b+b)}} = \sqrt{b^2 + a \sqrt{b^2 +(a+b)(a+2b+b)}}$ Applying the formula again to $a+2b+b$ gives $\sqrt{b^2 + a \sqrt{b^2 +(a+b) \sqrt{b^2 + (a+2b)(a+2b+b+b)}}} = \sqrt{b^2 + a \sqrt{b^2 +(a+b) \sqrt{b^2 + (a+2b)(a+3b+b)}}}$ Continuing in the fashion, we get: $a+b = \sqrt{b^2 + a \sqrt{b^2 +(a+b) \sqrt{b^2 + (a+2b) \sqrt{b^2 + (a+3b) \sqrt{b^2 + (a+4b) \sqrt{b^2 + \cdots}}}}}}$ This can be proven by induction. Putting $a = 4$ and $b = 1$ yields $5 = \sqrt{1 + 4 \sqrt{1 + 5 \sqrt{1 + 6 \sqrt{1 + \cdots}}}}$

A different approach:

Define $x_k = k \sqrt{1 + x_{k+1}}$ . We want $x_3/3$ . Let $M>3$ an integer, and define $y_j$ by $y_{M-k} = x_k$ . Then $y_j = (M - j) \sqrt{1+y_{j-1}}, y_0 = x_M$ . We now want $y_{M-3}/3$ . Note that for any $y_0 > -1$ , $y_{M} \rightarrow 15$ as $M \rightarrow \infty$ . So the answer is $15/3 = 5$ .

The equation is in the form $f(x)=\sqrt{1+x(f(x+1))}$ where $x=4$ . Note that $f(x)=x+1$ satisfies this equation, i.e. $x+1=\sqrt{1+x(x+2)}$ . Therefore, $n=f(4)=5$ .

Trivial by Desmos Theorem

Used WolframAlpha with increasing terms which seemingly are converging to (n+1) (here n=4 is the first term), therefore, 5 =4+1 is the answer.

Answer=5

However, it is interesting to show that solving

f(x)=√(1+n.f(1+x))

by assuming f(x)=x and f(1+x) = (1+x) gives simple solution

x=n+1, and

Answer = 5, ( n=4)

5=(1+24)^1/2 24=4(6)=4(1+35)^1/2 35=5(7)=5(1+48)^1/2 48=6(8)=6(1+63)^1/2 ,,,, countinouosly so n=5