Question

what are ramanujans infinte roots?

Answer 1

Answer:

Ramanujan has many infinite expression one of which is$3=\sqrt9 \\ 3=\sqrt{1+8} \\ 3=\sqrt{1+2\cdot4}  \\ 3=\sqrt{1+2\cdot\sqrt{16}} \\ 3=\sqrt{1+2\cdot \sqrt{1+15}} \\ 3=\sqrt{1+2\cdot\sqrt{1+3\cdot5} } \\ 3=\sqrt{1+2\cdot\sqrt{1+3\cdot\sqrt{25}} } \\ 3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+24}} } \\ 3=\sqrt{1+2\cdot \sqrt{1+3 \cdot \sqrt{1+4\cdot 6}} } \\ 3=\sqrt{1+2\cdot \sqrt{1+3\cdot \sqrt{1+4\cdot \sqrt{36}}} } \\so,\\ 3=\sqrt{1+2\cdot\sqrt{1+3\cdot\sqrt{1+4\cdot \sqrt{1+ \cdots}}} }$

Which can be generalised as

$n=\sqrt(n^2)  \\ n=\sqrt{1+n^2-1^2}  \\ n=\sqrt{1+(n-1) \cdot  (n+1)} \\   n=\sqrt{1+(n-1)\cdot \sqrt{(n+1)^2}}  \\ n=\sqrt{1+(n-1)\cdot \sqrt{1+(n+1)^2-1^2}} \\ n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot (n+2)} }   \\ n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{(n+2)^2}} }  \\ n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{1+(n+2)^2-1^2}} } \\ n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{1+(n+1)\cdot (n+3)}} }  \\ n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{1+(n+1)\cdot \sqrt{(n+3)^2}}} }$

So,

$n=\sqrt{1+(n-1)\cdot \sqrt{1+n \cdot \sqrt{1+(n+1)\cdot \sqrt{1+(n+2)}}} }.....$

Thus using the above expression we can turn any number greater than 1 to this infinte expression.

THANK YOU:)))

Answer 2

Answer:

it's too long bhro and sis but I am trying to get the correct answer