Find square roots upto infinte times

Question

Evaluate : $\sqrt{1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}}$

Is it possible to solve in the following way : Let
$x=\sqrt{1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}}$
$x^2= 1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}$

next ??????

I don't know

Answer 1

This question was given by Ramanujan. To solve this observe that $3=\sqrt{1+ 2 *4}$ and $4=\sqrt{1+ 3 *5}$ and so on... Now put the expression for 4 instead 4 appearing in RHS side of the expression for 3. Then put expression for 5 and so on. Convergence is the only thing to be proven. I is clear since it is bounded and also monotone.

Answer 2

Note that $(x+1)^2=1+x(x+2)$, now taking the principal square root of both sides and re-substituting the resulting expression for $x+1$ multiple times gives us that:

$$(x+1)=\sqrt{1+x(x+2)}$$ $$(x+1)=\sqrt{1+x\sqrt{1+(x+1)(x+3)})}$$ $$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}$$ $$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{1+(x+4)....}}}}}$$

$$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7}}...}}}}$$

You can use the monotone convergence theorem to prove that this resulting expression converges.