The sequence {a

_{n}} is defined by
a

_{1} = 1 and a

_{n+1} = a

_{n}+1/(a

_{n}^{2}).

Show that a_{2016} 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**