What is the exact value of the series $\sqrt{1+2\times\sqrt{1+3\times\sqrt{1+4\times\sqrt\cdot.....}}}$ Kindly tell me
what is the value of:
$\sqrt{1+2\times\sqrt{1+3\times\sqrt{1+4\times\sqrt\cdot.....}}}$
According to ramanujan ,it is equal to 3
I want to know how...
 A: This formula makes no sense. When u write "$\ldots$" it's like  saying "and so on" where you expect the reader to understand from context what you are talking about.
For example if commonly when u see $1+2+3+4+\ldots$ you would think about a limit of sequence of partial sums therefore you would said $1+2+3+4+\ldots=\infty$ since that sequence does not converge.
However, speaking of ramanujan, he used different context where he stated that $1+2+3+4+\ldots=-{1 \over 12}$ simply because he meant something different by "$\ldots$"
Returning to your question. Your "$\ldots$" does not point to any scheme.
ramanujan wrote something like it's 3 because
$3=\sqrt{9}=\sqrt{1+2\cdot4}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\cdot5}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}=\ldots=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}}$
Unfortunately such argument would work for any positive value.
For example $10=\sqrt{100}=\sqrt{1+2\cdot{99 \over 2}}=\sqrt{1+2\sqrt{1+3\cdot{                              9797 \over 12}}}=\ldots$
the weird fraction will become irrelevant
