While I was surfing on the Internet yesterday, I watched a video about Ramajuan's infinite root. After that I had tried on my own and I got the point.

$ 3=\sqrt9$

$3=\sqrt{1+8}$

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

$3=\sqrt{1+2\cdot \sqrt{16}}$

$3=\sqrt{1+2\cdot \sqrt{1+15}}$

$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot 5} }$

$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{25}} }$

$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+24}} }$

$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot 6}} }$

$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{36}}} }$

$\vdots$

$3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$

and I also tried

$ 4=\sqrt{16}$

$4=\sqrt{1+15}$

$4=\sqrt{1+2 \cdot \frac{15}{2}}$

$4=\sqrt{1+2\cdot \sqrt{\frac{225}{4}}}$

$4=\sqrt{1+2\cdot \sqrt{1+\frac{221}{4}}}$

$4=\sqrt{1+2\cdot \sqrt{1+ 3\cdot \frac{221}{12}}}$

$4=\sqrt{1+2\cdot \sqrt{1+ 3\cdot \sqrt{\frac{48841}{144}}}}$

$\vdots$

$4=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$

for the value 4. Isn't there a contradiction? I also tried with starting $2=\sqrt4$ and it does not work for infinetely times . I sensed that I can do same actions for every number which is greater than and equals to 3. For every help and opinion, thanks in advance :)