The sequence {a
n} is defined by
a
1 = 1 and a
n+1 = a
n+1/(a
n2).
Show that a2016 is over 18.
Lemma:
If a(n) > k-1, then a(n+k^2) > k
Proof by contradiction
The a(n) terms increase steadily, but by a smaller and smaller amount.
Assume a(n+k^2) < k
Then each of the k^2 terms between a(n+1) and a(n+k^2) (inclusive) is at least 1/k^2 greater than the previous term.
But this means that a(n+k^2) is at least k^2 * (1/k^2) greater than a(n).
Then, a(n+k^2) > 1+ a(n) > k
This is a contradiction, so the initial assumption is false, and the lemma is true.
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Thus, it takes less than 1 + 4 + 9 + ... + 49 + 64 = 204 terms to reach 8.
since a(204) > 8, so is a(2016).
Note that this proof uses a conservative estimate.
In fact, according to excel, a(169) = 8.0006639 is the first term that is greater than 8.
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The sum of the first n squares = n(n+1)(2n+1)/6,
so a[n(n+1)(2n+1)/6] > n.
In particular, a(1785) > 17.
So a(2016) > 17.
And again, it is conservative.
In fact, a(2016) = 18.22831144
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OOPS. NEVER MIND. I thought I was proving that a(2016) > 8. I guess I need to refine my methods
Edited on March 29, 2017, 4:01 pm
Conservatively speaking (spoiler)
