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Elsewhere we proved that the sequence

(1)	$\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}}\ldots$

is convergent to 2. With this in mind, it is possible and meaningful to shorten the description of this fact to

$2 = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots}}}}}$

Expressions like that serve an example of infinite nested radicals. They are realized in terms of limits, if those exist, and are understood as the shorthand for the latter.

For a real s > 0,

(2)	$\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}}$

always exists. For s < 2, this follows immediately from the Monotone Convergence theorem because then sequence (2) is bounded by the sequence (1) termwise. For s > 2, the derivation is a little more complex.

The curious observation about (2) is that, like in (1), the infinite nested radical may equal an integer. In general, let

$S = \sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}}$

Then, adding s to both sides,

$S + s = s + \sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}} = S^2.$

Thus we obtain a relationship between s and S: $s = S^{2}-S$ and using the quadratic formula also $S = (1 \pm \sqrt{1+4s})/2.$ The negative number ought to be recognized as spurious and discarded. So, finally we have

$\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}} = (1+\sqrt{1+4s})/2.$

on the one hand and

(3)	$S = \sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s+\sqrt{s\ldots}}}}}, for s = S^{2}-S,$

on the other. From (3) we get several examples:

$3 = \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6\ldots}}}}},$ $4 = \sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12\ldots}}}}},$ $5 = \sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20\ldots}}}}},$ $6 = \sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30+\sqrt{30\ldots}}}}},$

and so on.

An interesting quandary arises when we pick $S = 1$ which would require $s = 0$ and thus appear to imply

$1 = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0\ldots}}}}}$

which, obviously could not be right. Indeed, the right-hand side represents the limit of the sequence

$\sqrt{0},\sqrt{0+\sqrt{0}},\sqrt{0+\sqrt{0+\sqrt{0}}}\ldots$

with all terms zero. The sequence converges to 0, not 1! How does one explain that?

Answer →

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

How do you explain

$1 = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0\ldots}}}}}?$

At one point we derived the relation $S = (1 \pm \sqrt{1+4s})/2$ and dismissed one solution with the sign minus as being negative and therefore unsuitable. However, for s = 0, the two solutions are $(1 \pm 1)/2,$ i.e., 1 and 0. In this case, 1 is obviously spurious and should be dropped, leaving the correct identity

$0 = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0\ldots}}}}}.$

References

1. C. A. Pickover, A Passion for Mathematics, John Wiley & Sons, 2005, p. 96
2. C.C. Clawson, Mathematical Mysteries, Plenum Press, 1996, pp. 140-144