(In reply to
But... by Kenny M)
If a(n+1)=1+n/(a(n)) (The starting point is not that relevant for this purpose)
Then a(n + 1) = (a(n) + n)/a(n)
So a(n + 1)*a(n) = (a(n) + n)
Since a(n + 1) approximates to a(n) for large n, this is essentially the quadratic a(n)^2-a(n)-n = 0.
Edited on October 29, 2020, 9:11 pm
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Posted by broll
on 2020-10-29 21:05:39 |