About (One of the first problems of Ramanujan) $\sqrt{1+2\sqrt{1+3\sqrt{1+4....}}}$

One of the first problems Ramanujan posed in the journal was to find the value of: $\sqrt{1+2\sqrt{1+3\sqrt{1+4....}}}$

He waited for a solution to be offered in three issues, over six months, but failed to receive any. At the end, Ramanujan supplied the solution to the problem himself. On page 105 of his first notebook, he formulated an equation that could be used to solve the infinitely nested radicals problem. $x+n+a=\sqrt{ax+(n+a)^2+2\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{..}}}$

Using this equation, the answer to the question posed in the Journal was simply 3, obtained by setting x = 2, n = 1, and a = 0.

click here for to get info about Ramanujan. this information is shared from Srinivasa ramanujan wikipedia

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

About (One of the first problems of Ramanujan)1+21+31+4....\sqrt{1+2\sqrt{1+3\sqrt{1+4....}}}1+21+31+4....​​​

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About (One of the first problems of Ramanujan) $\sqrt{1+2\sqrt{1+3\sqrt{1+4....}}}$