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The basic idea is to take advantage of the identity:
(x + k a)² = (a² + (x + (k -1) a) (x + k a + a))
So, starting with k = 1, we have:
(x + a) = √(a² + x(x + a + a)))
which leads to:
(x + a) = √(a² + x√(a² + (x + a)(x + 2a + a))
and so on, so that we have:
(x + a) = √(a² + x√(a² + (x + a)√(a² + (x + 2a)√(a² + (x + 3a)√(a² + ...
Letting x = 2 and a = 1, we have:
2 + 1 = 3 = √( 1 + 2√(1 + 3√(1 + 4√(1 + ...
so there
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x + n + a = a x + ( n + a ) 2 + x a ( x + n ) + ( n + a ) 2 + ( x + n ) a ( x + 2 n ) + ( n + a ) 2 + ( x + 2 n ) ⋯ The above formula was discovered by Ramanujan.If we let a = 0 , n = 1 ,after simplifying we get: x + 1 = 1 + x 1 + ( x + 1 ) 1 + ( x + 2 ) ⋯ Our question fits this form,so letting x = 2 ,we get: 2 + 1 = 3 = 1 + 2 1 + 3 1 + 4 ⋯