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The Largest term in the Sequence \(\displaystyle \frac{1}{503}\;,\frac{4}{524}\;,\frac{9}{581}\;,\frac{16}{692},.....\) is

Good morning,

The main difficulty in this problem was to find the general term of the sequence. Especially in relation to the denominator. After some effort we have the following:

\(a_n = \frac{n^2}{500 + 3n^3}\)

Clearly, when \(n\) tends to infinity then \(a_n\) tends to \(0\) so \((a_n)\) have a largest term.

So we must derivate \(a_n\):

\(a'_n = \frac{2n(500+3n^3)-n^2(9n^2)}{(500+3n^3)2}\)

Now we set this result to zero to find \(n\):

\(a'_n = \frac{2n(500+3n^3)-n^2(9n^2)}{(500+3n^3)2} = 0 <=> 2n(500+3n^3)-n^2(9n^2) = 0 <=> 2n(500+3n^3) = n^2(9n^2)\)

So \(1000 + 6n^3 = 9n^3 <=> n^3 = \frac{1000}{3} <=> n \approx 6,9\), which leads to \(n = 7\) and \(a_7 = \frac{7^2}{(500+3\cdot7^3)}\) is the largest term in the sequence.

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Fraol
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Hi,

To verify the above results, considering that the general term formula is correct, I tabulated the first 15 terms of the sequence:


.

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Fraol
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