Ramanujuan

Prove that \(3=\sqrt{1+2\sqrt{1+3 \sqrt{1+4 \sqrt{1+ \cdots}}}}\).

Source: https://www.quora.com/What-is-the-solution-of-sqrt-1+2-sqrt-1+3-sqrt-1+4-sqrt

\(\small{ \begin{array}{|rcll|} \hline n(n+2) &=& n\sqrt{(n+2)^2} \\ &=& n\sqrt{n^2+4n+4} \\ &=& n\sqrt{1+n^2+4n+3} \\ &=& n\sqrt{1+n^2+3n+n+3} \\ \mathbf{ n(n+2) } &=& \mathbf{ n\sqrt{1+(n+1)(n+3)} } \\ \hline f(n) &=& n(n+2) \\\\ f(n+1) &=& (n+1)(n+1+2 ) \\ f(n+1) &=& (n+1)(n+3) \\ \mathbf{ n(n+2) } &=& \mathbf{ n\sqrt{1+(n+1)(n+3)} } \\ f(n) &=& n\sqrt{1+f(n+1)} \\ && \boxed{f(n+1) = (n+1)\sqrt{1+f(n+1+1)}\\ f(n+1) = (n+1)\sqrt{1+f(n+2)} } \\ f(n) &=& n\sqrt{1+(n+1)\sqrt{1+f(n+2)}} \\ && \boxed{f(n+2) = (n+2)\sqrt{1+f(n+2+1)}\\ f(n+2) = (n+2)\sqrt{1+f(n+3)} } \\ f(n) &=& n\sqrt{1+(n+1)\sqrt{1+ (n+2)\sqrt{1+f(n+3)}}} \\ && \boxed{f(n+3) = (n+3)\sqrt{1+f(n+3+1)}\\ f(n+3) = (n+3)\sqrt{1+f(n+4)} } \\ f(n) &=& n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+f(n+4)}}}} \quad | \quad f(n) = n(n+2)\\ n(n+2) &=& n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+f(n+4)}}}} \quad | \quad n = 1 \\ 1(1+2) &=& 1*\sqrt{1+(1+1)\sqrt{1+(1+2)\sqrt{1+(1+3)\sqrt{1+\dots}}}} \\ 3 &=& \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\dots}}}} \\ \hline \end{array} }\)

Wow! That sure is an interesting question. I plugged the first numbers into my calculator and it came close to 3, so I think it is correct, but "It works because my calculator said so" isn't a good proof (unfortunately).

I do know that we can solve for x when we have: \(x = \sqrt{y+\sqrt{y+\sqrt{y+...}}}\) because we can square both sides, subtract by y, and then the RHS of the equation is x, so we are left with \(x^{2}-y=x\).

However, there is no reoccurring pattern in your problem, it is just an ongoing arithmetic sequence. So I'm not really not sure how to find it. However, if you set the RHS equal to x, square both sides of your equation, and then play with it to get the RHS of your equation to come up again, then you can solve it.