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Theorems for Differentiability

Let a n , b...

Question

Let ${a_{n}}, {b_{n}}, {c_{n}}$ be sequences such that
$(i) a_{n} + b_{n} {+ c}_{n} = 2 n + 1$
$(i i) a_{n} b_{n} + b_{n} c_{n} + {+ c}_{n} a_{n} = 2 n - 1$
$(i i i) a_{n} b_{n} c_{n} = - 1$
$(i v) a_{n} < b_{n} < c_{n}$
Then find the value of $lim n \to \infty n a_{n}$ .

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Solution

$a_{n} + b_{n} + c_{n} = (2 n + 1) (b_{n} + c_{n}) = (2 n + 1) - a_{n} (1) a_{n} b_{n} + b_{n} c_{n} + c_{n} a_{n} = (2 n - 1) a_{n} (b_{n} + c_{n}) + b_{n} c_{n} = (2 n - 1) (2) a_{n} b_{n} c_{n} = - 1 b_{n} c_{n} = \frac{- 1}{a_{n}} (3)$
From $(1), (2)$ and $(3)$
$a_{n} ((2 n + 1) - a_{n}) - \frac{1}{a_{n}} = (2 n - 1) {a_{n}}^{3} - (2 n + 1) {a_{n}}^{2} + (2 n - 1) a_{n} + 1 = 0$
We can check that $(a_{n} = 1)$ is a root of this equation
$(1)^{3} - (2 n + 1) + (2 n - 1) + 1 = 0$

Dividing the equation be $(a_{n} - 1)$ , we get ${a_{n}}^{2} - 2 n a_{n} - 1$
${a_{n}}^{3} - (2 n + 1) {a_{n}}^{2} + (2 n - 1) a_{n} + 1 = 0 (a_{n} - 1) ({a_{n}}^{2} - 2 n a_{n} - 1) = 0 (a_{n} - 1) ({a_{n}}^{2} - 2 n a_{n} + n^{2} - n^{2} - 1) = 0 (a_{n} - 1) [(a_{n} - n)^{2} - (n^{2} + 1)] = 0 (a_{n} - n)^{2} - (n^{2} + 1) = 0 (a_{n} - n)^{2} = (n^{2} + 1) (a_{n} + n) = \pm \sqrt{n^{2} + 1} a_{n} = - n \pm \sqrt{n^{2} + 1} a_{n} = 1, - n + \sqrt{n^{2} + 1}, - n - \sqrt{n^{2} + 1} lim n \to \infty n a_{n} = \infty$

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