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