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 :)