# Show that $3=\sqrt{1+2\sqrt{1+3\sqrt{1+4…}}}$ [duplicate]

I was solving a problem, asking me to prove the identity $3=\sqrt{1+2\sqrt{1+3\sqrt{1+4...}}}$ which was first posed by Ramanujan.
The standard answer goes as $$3=\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{1+3\cdot5}}=\dots=\sqrt{1+2\sqrt{1+3\sqrt{1+4...}}}$$ which I think can be stated rigorously as "Let $$x_n=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\dots\sqrt{1+(n\cdot n+2)}}}}}$$ then $x_n=3$ for every $n$."
However, I interpret this problem to be "Let $y_n=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\dots\sqrt{1+n}}}}}$, then $\lim_{n\to\infty}y_n=3$."
It is obvious that $y_n\le3$ and is increasing, therefore $\lim_{n\to\infty}y_n\le3$. I believe that they are actually equal, but have trouble estimating the difference $|3-y_n|$. Does anyone have any idea?