# CTK Insights

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02 Aug

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?

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

#### 4 Responses to “Nested radicals”

1. 1
Xamuel Says:

Nice trick there with the spurious 1. I feel like this well deserves a place with all the other famous "1=0 contradictions" where the student is required to find the flaw.

2. 2

Yes, I agree.

In addition, this is a sort of interplay between algebra and analysis that is often disregarded, e.g., in questioning whether .999... = 1 or not. Several math notations include ellipses and have no meaning, except as limits. Once this is understood, algebra may be be fruitfully employed.

3. 3
Hubert Shutrick Says:

The more obvious way to get S^2 = s + S is to just square the equation.

4. 4