Evaluate $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$ $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
My Attempt: I tried to use the regular way.
$A=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$
$A^2=1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}$
But then I saw that nothing happened twice and I stopped. Any hints?
It's better to don't use limits but if you used no problem. If there is an answer that doesn't use limits will accept.
update1: The question that look likes this  wants to find the limit but I want a much easier way to solve it. But if there isn't any answer easier answer the limition one will accept.
update2: You solve the problem when you know the answer is 3 think that you don't know that the answer is 3 then solve.

First you can guess that this seems to converge to $3$ by using a calculator (or by estimating the radicals). Assuming the limit is three, then one can try to wrtie $3$ as the following down:
$3=\sqrt{1+8}=\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+15}}=\sqrt{1+2\sqrt{1+3*5}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot 6}}}=\ldots$
This iteration is the same as the formula you stated above, because it holds
$\sqrt{n^2}=\sqrt{1+(n-1)(n+1)}=\sqrt{1+(n-1)\sqrt{(n+1)^2}} \quad \forall n\in \mathbb{N}, n>1$
Thus, the limit is $3$.
