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!
As e.g. points out, it seems to be equal to 3.
In addition, starting farther along in the sequence yields interesting results. For example:
2(1+ 3(1+ 4(1+ 5(...)))) --------> 8
3(1+ 4(1+ 5(1+ 6(...)))) --------> 15
.
.
.
n(1+ (n+1)(1+ (n+2)(1+ (n+3)(...)))) --------> n^2 - 1
and also
(1+ 3(1+ 4(1+ 5(...)))) --------> 4
(1+ 4(1+ 5(1+ 6(...)))) --------> 5
(1+ 5(1+ 6(1+ 7(...)))) --------> 6
etc...
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Posted by Larry
on 2004-10-01 04:06:02 |
interesting
