Answer:

First of all, thank you for asking such a beautiful question. I am a fan of Ramanujan.

In Ramanujan’s 2nd Notebook, Chapter XII, Page 108.

Ramanujan, being the genius he was, solve the problem in the most elegant of ways.

I will put it up here for the sake of completeness,

n(n+2)=n1+(n+1)(n+3)−−−−−−−−−−−−−−−√

Let,

f(n)=n(n+2)

f(n)=n1+(n+1)(n+3)−−−−−−−−−−−−−−−√

f(n)=n(1+f(n+1))−−−−−−−−−−−√

f(n)=n1+(n+1)1+(n+2)1+(n+3)−−−−−−−−−√−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

f(n)=…

That is,

n(n+2)=n1+(n+1)1+(n+2)1+(n+3)1+…−−−−−−√−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

Putting, n=1 we have,

1+21+31+41+…−−−−−−√−−−−−−−−−−−√−−−−−−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−−−−−−√=3

Vijayaraghavan proved that a sufficient condition

for the convergence of the following sequence

a1+a2+…+an−−√−−−−−−−−−−−−√−−−−−−−−−−−−−−−−−−√

is that,

limn→∞¯logan2n<∞

ANOTHER FAMOUS METHOD IS, (reverse engineering).

3=9–√=1+8−−−−√=1+2⋅4−−−−−−−√

1+2⋅4−−−−−−−√=1+2⋅16−−√−−−−−−−−−√=1+2⋅1+3⋅5−−−−−−−√−−−−−−−−−−−−−√

Keep going in that way and get the given radical.

Thank you. Cheers! ⌣¨

References:

Step-by-step explanation:

Given that:-