Amazing Nested Radical 2
Compute:
3 4 + 1 2 3 1 0 + 3 2 3 1 6 + 5 2 3 2 2 + 7 2 3 2 8 + ⋯
Details and assumptions:
The first sequence 4 , 1 0 , 1 6 , 2 2 , … is an arithmetic progression , while the second one 1 2 , 3 2 , 5 2 , 7 2 , … is the sequence of odd squares.
The answer is 2.
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1 solution
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This is actually a slight generalization of well-known Ramanujan's identity: 1 + 2 1 + 3 1 + 4 ⋯ = 3
_ PROOF SKETCH (Ramanujan-like argument): _ x + 1 = 3 ( x + 1 ) 3 = 3 1 + 3 x + x 2 ( x + 3 ) =
3 1 + 3 x + x 2 3 ( x + 3 ) 3 = 3 1 + 3 x + x 2 3 7 + 3 x + ( x + 2 ) 2 ( x + 5 ) . . .
and all that remains is to plug in x = 1 .
Of course, this is not rigorous, but it shows and explains the main idea.