hihi, all - since i didn't know what to do with this question, i wrote a program to generate some values; i assumed that if f(n) is really large, then the term in parentheses is only a little more than 1, so that f(n+1) may not be very large (depending on n, so a kind of slow divergence might be possible) it turned out much weirder than that - there appears to be a threshold t ~ 1.84 (i have more digits) with the property that if f(1) > t, then the sequence eventually alternates f(n) = 1 for odd n, f(n) = 2 ^ n for even n, and if f(1) < t, then the sequence eventually alternates f(n) = 1 for even n, f(n) = 2 ^ n for odd n the reason this is only partly a spoiler is that this depends on finite precision computation, so that when n is large enough, 1 + 1/2^n = 1, and the rest follows easily, and of course that means that this might be completely bogus for the actual sequence but this might stimulate some thought (and the threshold seems strange) more later, chris On 2020-08-30 19:12, Dan Asimov wrote:
I found this interesting question in an undisclosed location:
Let f : Z+ —> R be a function such that
a) f(1) > 0
and
b) f(n+1) = (1 + 1/f(n))^n
for all n >= 1.
Is it possible that f(n) —> oo as n —> oo ???
—Dan
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