Let-lt-x-n-gt-be-a-sequence-defined-by-x-n-1-1-k-x-n-k-x-n-n-N-Show-that-lt-x-n-gt-converges-to-k-k-1-x-1-gt-0-k-gt-1-

Question Number 141525 by hgrocks last updated on 20/May/21

Let < x_{n} > be a sequence defined by

x_{n + 1} = \frac{1}{k} (x_{n} + \frac{k}{x_{n}}) \forall n \in N

Show that < x_{n} > converges to \sqrt{\frac{k}{k - 1}}

x_{1} > 0, k > 1

Answered by mathmax by abdo last updated on 21/May/21

x_{n + 1} = f (x_{n}) with f (x) = \frac{1}{k} (x + \frac{k}{x}) f is continue on R^{★}

f^{^{'}} (x) = \frac{1}{k} (1 - \frac{k}{x^{2}}) = \frac{1}{k} \times \frac{x^{2} - k}{x^{2}} = \frac{1}{{kx}^{2}} (x - \sqrt{k}) (x + \sqrt{k})

we can limit the variation on] 0, + \infty [due to u_{n} > 0 for all n

x 0 \sqrt{k} + \infty

f^{^{'}} ∣ ∣ - 0 +

f ∣ ∣ dec f (\sqrt{k}) incr + \infty

due to continuity of f the limit of u_{n} is the fix point of f

f (x) = x \Rightarrow \frac{1}{k} (x + \frac{k}{x}) = x \Rightarrow x^{2} + k = {kx}^{2} \Rightarrow (k - 1) x^{2} = k \Rightarrow

but x > 0 \Rightarrow x = \sqrt{\frac{k}{k - 1}} (k > 1)

Commented by hgrocks last updated on 21/May/21

Thanks a lot . ★ ★

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