GROCERY & COUPON CODES

v[{ (1 + 2v( 1 + 3v( 1 + 4v ( 1 + 5v( 1 +....))))}] = P?

If P is to have any meaning then it must be as the limit of an infinite seuqence of terms. But how do we produce those terms by truncating the given expression? If you agree to truncate in front of each square root sign then the sequence is
sqrt(1 + 2)
sqrt(1 + 2*sqrt(1 + 3))
sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4)))
sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(1 + 5))))
etc.
Evaluating the first 9 terms to 3 decimal places (except the last 2) gives
1.732, 2,236, 2.560, 2.755, 2.867, 2.949, 2.963, 2.98953, 2.99462
This suggests that the sequence is heading for a limit of 3.
EDIT. Let us suppose that the limit is 3.
Also suppose that the limit of everything after the 2 is Q. Then we have
sqrt(1 + 2Q) = 3 ----> 1 + 2Q = 9 ----> Q = 4
Now suppose that everything to the right of the 3 has a limit of R. Then we have
sqrt(1 + 2*sqrt(1 + 3R)) = 3
1 + 2*sqrt(1 + 3R) = 9
sqrt(1 + 3R) = 4
1 + 3R = 16
R = 5
P = 3, Q = 4, R = 5, looks like it makes sense in some way.
EDIT 2. I've finally managed to work out a sort of (altered) proof P = 3.
Assume that
sqrt(1 + a*sqrt(1 + (a + 1)*sqrt(1 + (a + 2)*sqrt ....... )))) = a + b
Square both sides
1 + a*sqrt(1 + (a + 1)*sqrt(1 + (a + 2)*sqrt ...... )))) = a^2 + 2ab + b^2
sqrt(1 + (a + 1)*sqrt(1 + (a + 2)*sqrt(1 + ..... )))) = a + 2 + (b^2 - 1)/a
Replace a by a - 1
sqrt(1 + a*sqrt(1 + (a + 1)*sqrt(1 + (a + 2)*sqrt ..... ))) = a + 1 + (b^2 - 1)/(a - 1)
Since the left is the original line then so must be the right, so
a + 1 + (b^2 - 1)/(a - 1) = a + b
a^2 - 1 + b^2 - 1 = a^2 + ab - a - b
b^2 - ab + a + b - 2 = 0
(b - 1)(b - a + 2) = 0
b = 1 or a - 2
P = a + 1 or 2a - 2
When a = 2 this gives P = 3 or 2. Not easy to explain why there are two results.
However, a = 0 shows that P = 2a - 2 is a false result.
EDIT 3. I have belatedly realised that what is written above, if re-ordered, constitutes a proof by induction. To prove
sqrt(1 + n*sqrt(1 + (n + 1)*sqrt(1 + . . . ))) = n + 1
First note that it is certainly true for n = 0 as both sides are 1.
Then assume true for n and
sqrt(1 + (n + 1)*sqrt(1 + (n + 2)* . . . ))) = [(n + 1)^2 - 1]/n = n + 2
Thus if true for n it is also true for n + 1. Hence true for all n by induction.

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