How much is 1√(1+ 2√(1+ 3√(1+ 4√(1+ 5√(...)))))? Does it tend to infinity?
Note: the initial 1 is just for completeness!
(In reply to
re: interesting by Ira Jewell)
you cannot divide the RHS by n when n is zero so the inductive step doesn't work for n=0. You'll see why this division is necessary for the step if you actually do it.
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Posted by Bon
on 2004-10-06 03:42:48 |