Inspiration: Ramanujan

According to Ramanujan (Equation (27) in Nested Radical ):

$x + 1 = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+...}}}}$

For $\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+...}}}}$ , implies that $x=1$ , therefore,

$\sqrt{1+\sqrt{1+2\sqrt{1+3\sqrt{1+...}}}} = 1 + 1 = \boxed{2}$

This is an interesting problem. Visit http://mathworld.wolfram.com/NestedRadical.html for a detailed solution. $\sqrt {1+ \sqrt { 1+2\sqrt { 1+3\sqrt { 1+4\sqrt { 1+5\sqrt{\cdots} } } } } }=\sqrt{1+3}=2$

$\sqrt { 1 + \sqrt { 1 + 2 \sqrt { 1 + 3 \sqrt { 1 + 4 \sqrt { 1 + 5 \sqrt { \cdots } } } } } } = \sqrt { 1 + \sqrt { 3 \sqrt { 4 \sqrt { 5 \sqrt { 6 \sqrt { \cdots } } } } } } = \boxed { 2 }$

We note that:

So

We get

We substitute the value of (n+3) by putting l=3 in the first equation:

And so on:

Then we get the result, putting n=1:

Therefore