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```
Date: 02/25/2002 at 19:58:50
From: Mac

sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+...)))) = 3

Thank you very much.
```

```
Date: 02/25/2002 at 21:01:47
From: Doctor Wilkinson

Nice problem, Mac! Here's the solution given by Ramanujan. You'll need
to fuss a bit about convergence.

Consider the identity

[x+n]^2 = n^2 + 2nx + x^2 = n^2 + x [(x+n) + n]

Take square roots to get

[x+n] = sqrt(n^2 + x [(x+n)+n])

Now inside the brackets you have something + n, so you can substitute
in the same equation with x+n replacing x to get

x+n = sqrt(n^2 + x sqrt(n^2 + (x+n)[(x+2n)+n]))

and repeat the process to get

x+n = sqrt(n^2 + x sqrt(n^2 + (x+n) sqrt(n^2 + (x+2n) sqrt(...)...)

If you now set n = 1, x = 2 you get

3 = sqrt(1 + 2 sqrt(1 + 3 sqrt(1 + 4 sqrt(...)...)

- Doctor Wilkinson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 02/25/2002 at 22:19:24
From: Mac

Thank you very much. I really appreciated your response.
```
Associated Topics:
High School Exponents
High School Sequences, Series

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