Nested radicals like Ramanujan's infinite radicals $$\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+...}}}}}=?$$
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My question is about "how to start this problem ?".I've been thinking for over two hours but get stuck at the end. I tried by a calculator to approximate the value, I find out if I define a sequence like below,It seems $a_n\to 2$ $$a_1=\sqrt1\\
a_2=\sqrt{1+\sqrt{5}}\sim 1.798907\\
a_3=\sqrt{1+\sqrt{5+\sqrt{11}}}\sim 1.97075\\
a_4=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19}}}}\sim 1.99661\\
a_5=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29}}}}}\sim 1.99967\\\vdots \\a_n  \text{tends to } 2$$
But I have no idea to solve it analytically. Can someone help me? or get me the clue?

$$2=\sqrt{1+\sqrt9}=\sqrt{1+\sqrt{5+\sqrt{16}}}=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{25}}}}=$$
$$=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{36}}}}}=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{49}}}}}}=...$$
$$=\sqrt{1+\sqrt{5+\sqrt{11+...+\sqrt{n^2+n-1+\sqrt{(n+2)^2}}}}}=$$
$$=\sqrt{1+\sqrt{5+\sqrt{11+...+\sqrt{n^2+n-1+\sqrt{(n+1)^2+(n+1)-1+\sqrt{(n+3)^2}}}}}}=...$$
