We are in the right year for this! - (3)

Calculus Level 5

lim ⁡ n → ∞ ( u n + 1 u 1 u 2 u 3 ⋯ u n ) 2 \large \lim_{n\to\infty} \left( \frac{u_{n+1}}{u_1 u_2 u_3\cdots u_n}\right)^2

Given a recurrence relation u n + 1 = u n 2 − 2 u_{n+1} = u_n ^2 - 2 with u 1 = 2015 u_1 = \sqrt{2015} , find the value of the limit above.


The answer is 2011.

Alisa Meier by Alisa Meier , 24, Germany

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Let us define θ ∈ R \theta\in \mathbb{R} , such that u 1 = 2 cos ⁡ h θ u_1=2\cos h\theta ; then, from the given recursion we deduce that u n = 2 cos ⁡ h 2 n θ u_n=2\cos h 2^n\theta . Then the given limit reduces to lim ⁡ n → ∞ ( u n + 1 ∏ k = 1 n u k ) 2 = ( lim ⁡ n → ∞ 2 cos ⁡ h 2 n θ ∏ k = 1 n ( 2 cos ⁡ h 2 k θ ) ) 2 = ( lim ⁡ n → ∞ 2 sin ⁡ h θ cot ⁡ h 2 n θ ) 2 = ( u 1 2 − 4 ) lim ⁡ n → ∞ cot ⁡ h 2 n θ = ( u 1 2 − 4 ) = 2011 \lim_{n\to \infty}\left(\frac{u_{n+1}}{\prod_{k=1}^n u_k}\right)^2\\=\left(\lim_{n\to \infty}\frac{2\cos h2^n\theta}{\prod_{k=1}^n(2\cos h 2^k \theta)}\right)^2\\=\left(\lim_{n\to \infty}2\sin h\theta \cot h 2^n\theta\right)^2\\=(u_1^2-4)\lim_{n\to \infty}\cot h 2^n\theta=(u_1^2-4)=\boxed{2011}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...