If \(\left( 1 + \sqrt{3} \right)^n = A + B\sqrt{3}\), show that the limit of \(\frac{A}{B}\) as \(n\) approaches infinity is equal to sqrt(3).

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TopNewestIt must be \(\sqrt{3}\) rather. Two ways to solve: \((1 + \sqrt{3})^n = \displaystyle \sum_{r = 0}^{n} {n \choose r}({\sqrt{3}})^r = A + B\sqrt{3} \)

\((1 - \sqrt{3})^n = \displaystyle \sum_{r = 0}^{n} {n \choose r}({-\sqrt{3}})^r = A - B\sqrt{3} \) (Check that \(B\) comes from odd powers and \(A\) from even)

Solve to get :

\(A = \frac{(1 + \sqrt{3})^n + (1 - \sqrt{3})^n}{2}\) ,

\(B = \frac{ (1 + \sqrt{3})^n - (1 - \sqrt{3})^n}{2\sqrt{3}}\)

Now, \(\displaystyle \lim_{n \to \infty} \frac{A}{B}\)

= \(\frac{\frac{(1 + \sqrt{3})^n + (1 - \sqrt{3})^n}{2}}{\frac{ (1 + \sqrt{3})^n - (1 - \sqrt{3})^n}{2\sqrt{3}}}\)

= \(\boxed{\sqrt{3}}\) . \(\Bigg(As \displaystyle \lim_{n \to \infty} \frac{(1 - \sqrt{3})^n}{(1 + \sqrt{3})^n} = 0 \Bigg)\)

\[ALTERNATE:\]

\(\displaystyle \lim_{n \to \infty} (1 - \sqrt{3})^n = \displaystyle \lim_{n \to \infty} A - B\sqrt{3} = 0 \) [As \(\displaystyle \lim_{n \to \infty} (1 - \sqrt{3})^n = 0\)]

\(\Rightarrow \displaystyle \lim_{n \to \infty} \frac{A}{B} = \boxed{\sqrt{3}}\)

This was less a limit than a binomial theorem question :) – Jatin Yadav · 3 years, 10 months ago

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Nice solution – Jagdish Singh · 3 years, 10 months ago

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Strangely, this reminds me of Pell's Equations. – Matthew Fan · 3 years, 10 months ago

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– Mark Hennings · 3 years, 10 months ago

Not surprising. The integer solutions of the equation \[ x^2 - 2y^2 \; = \; \pm1 \] are precisely the integer coefficients \(x + y\sqrt{2} = \pm(1 + \sqrt{2})^n\) for \(n \in \mathbb{Z}\) (you get \(+1\) when \(n\) is even, and \(-1\) when \(n\) is odd). Similar games can be played with \(\sqrt{3}\).Log in to reply

I'm assuming you mean \(A, B, n\) to be positive integers? Otherwise the problem is unsolvable. – Michael Tang · 3 years, 10 months ago

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Sometimes we even calculate \(\displaystyle \lim_{n \to \infty} f(n!) \dots \) . – Jatin Yadav · 3 years, 10 months ago

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