<s>[INST] Ramanujan's Nested Square Roots Well, while my friend was watching "The Man Who Knew Infinity"(Story of The Great Mathematician - THE Ramanujan), he came across the following equation ->
$3 = \sqrt{1+2{\sqrt{1+3\sqrt{1+4\sqrt{1+5.....\infty}}}}}$ (I hope you understand the series )
I actually found 1 method of proving this ->
We know that
$(x+1)^{2}=x^{2}+2x+1
          =1+x(x+2)$ -------------------> (1)
Substituting $2$ in the above identity
$(2+1)^{2}=1+2(2+2)$
$=>(3)^{2}=1+2(4)$
$=>3=\sqrt{1+2(4)} $ -------------------> (2)
Similarly , substituting $3$ in the above identity
$(3+1)^{2}=1+3(3+2)$
$=>(4)^{2}=1+3(5)$
$=>4=\sqrt{1+3(5)}$ -------------------> (3)
Similarly , substituting $4$ in the above identity
$(4+1)^{2}=1+4(4+2)$
$=>(5)^{2}=1+4(6)$
$=>5=\sqrt{1+4(6)}$ -------------------> (4)
Substituting value of (3) & (4) in (2)
$(2+1)^{2}=1+2(2+2)$
$=>(3)^{2}=1+2(4)$
$=>3=\sqrt{1+2(4)}$
$=>3=\sqrt{1+2\sqrt{1+3(5)}}$
$=>3=\sqrt{1+2\sqrt{1+3\sqrt{1+4(6)}}}$
We notice we can do this FOREVER, thus , arriving at
$3 = \sqrt{1+2{\sqrt{1+3\sqrt{1+4\sqrt{1+5.....\infty}}}}}$
Is there any other method to arrive at the given solution , in fact , PROVE the given equation ??
 [/INST] You can find it here. You basically find a generalization of the function, make it a recursive formula, and then solve it.
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