Trigonometric Recurrence and its Limit!

Geometry Level 5

x n + 1 = ( 2 + cos ⁡ ( 2 α ) ) x n + cos ⁡ 2 ( α ) ( 2 − 2 cos ⁡ ( 2 α ) ) x n + 2 − cos ⁡ ( 2 α ) \large{ x_{n+1} = \dfrac{(2 + \cos(2\alpha))x_n + \cos^2(\alpha)}{(2 - 2\cos(2\alpha))x_n + 2 - \cos(2\alpha)} }

Consider the sequence of real numbers ( x n ) n = 1 ∞ (x_n)_{n=1}^\infty defined as above with x 1 = 1 x_1 = 1 for every n ∈ N n \in \mathbb N , where α \alpha is a real parameter.

∀ n ∈ N \forall \, n \in \mathbb N , let y n = ∑ k = 1 n 1 2 x k + 1 \large{y_n = \displaystyle \sum_{k=1}^n \dfrac{1}{2x_k +1}}

For specific values of α \alpha , if the sequence ( y n ) n = 1 ∞ (y_n)_{n=1}^\infty has a finite limit, find that limit.

Bonus: Determine the values of α \alpha for which the sequence ( y n ) n = 1 ∞ (y_n)_{n=1}^\infty has a finite limit.


The answer is 0.500.

Satyajit Mohanty by Satyajit Mohanty , 24, India

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1 solution

Note that x n > 0 x_n>0 for all n n . We have:

2 x n + 1 + 1 = ( 4 + 2 cos ⁡ 2 α + 2 − 2 cos ⁡ 2 α ) x n + 2 cos ⁡ 2 α + 2 − cos ⁡ 2 α ( 2 − 2 cos ⁡ 2 α ) x n + 2 − cos ⁡ 2 α 2x_{n+1}+1=\dfrac{(4+2\cos2\alpha+2-2\cos2\alpha)x_n+2\cos^2\alpha+2-\cos2\alpha}{(2-2\cos2\alpha)x_n+2-\cos2\alpha}

= 3 ( 2 x n + 1 ) ( 2 − 2 cos ⁡ 2 α ) x n + 2 − cos ⁡ 2 α \qquad\qquad\;\,=\dfrac{3(2x_n+1)}{(2-2\cos2\alpha)x_n+2-\cos2\alpha}

which leads to

1 2 x n + 1 + 1 = ( 1 − cos ⁡ 2 α ) ( 2 x n + 1 ) + 1 3 ( 2 x n + 1 ) = 1 3 ( 2 sin ⁡ 2 α + 1 2 x n + 1 ) \dfrac{1}{2x_{n+1}+1}=\dfrac{(1-\cos2\alpha)(2x_n+1)+1}{3(2x_n+1)}=\dfrac{1}{3}\left(2\sin^2\alpha+\dfrac{1}{2x_n+1}\right) .

This show that: 1 2 x n + 1 + 1 − sin ⁡ 2 α = 1 3 ( 1 2 x n + 1 − sin ⁡ 2 α ) \dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\dfrac{1}{3}\left(\dfrac{1}{2x_n+1}-\sin^2\alpha\right) ,

Or z n + 1 = 1 3 z n z_{n+1}=\dfrac{1}{3}z_n , where z n = 1 2 x n + 1 − sin ⁡ 2 α z_n=\dfrac{1}{2x_n+1}-\sin^2\alpha .

Thus ( z n ) (z_n) is a geometric progression with z 1 = 1 3 − sin ⁡ 2 α z_1=\dfrac{1}{3}-\sin^2\alpha and ratio q = 1 3 q=\dfrac{1}{3} . Then

∑ k = 1 n z k = ( 1 3 − sin ⁡ 2 α ) 1 − 1 3 n 1 − 1 3 = 1 2 ( 1 − 3 sin ⁡ 2 α ) ( 1 − 1 3 n ) \displaystyle \sum_{k=1}^n z_k=\left(\dfrac{1}{3}-\sin^2\alpha\right)\dfrac{1-\dfrac{1}{3^n}}{1-\dfrac{1}{3}}=\dfrac{1}{2}\left(1-3\sin^2\alpha\right)\left(1-\dfrac{1}{3^n}\right)

Thus, y n = ∑ k = 1 n ( z k + sin ⁡ 2 α ) = 1 2 ( 1 − 3 sin ⁡ 2 α ) + n sin ⁡ 2 α \displaystyle y_n=\sum_{k=1}^n(z_k+\sin^2\alpha)=\dfrac{1}{2}\left(1-3\sin^2\alpha\right)+n\sin^2\alpha .

Since the sequence ( 1 3 n ) \left(\dfrac{1}{3^n}\right) converges, the sequence ( y n ) (y_n) converges if and only if the sequence n sin ⁡ 2 α n\sin^2\alpha converges, or s i n 2 α = 0 sin^2\alpha=0 , that is α = k π , k ∈ Z \alpha=k\pi,\; k\in\mathbb{Z} .

Hence, lim ⁡ n → ∞ y n = lim ⁡ n → ∞ 1 2 ( 1 − 1 3 n ) = 1 2 \displaystyle\mathop{\lim}\limits_{n\to\infty}y_n=\mathop{\lim}\limits_{n\to\infty}\dfrac{1}{2}\left(1-\dfrac{1}{3^n}\right)=\boxed{\dfrac{1}{2}} .

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