Question

[Dãy số]

Answer 1

[Lê Song Phương]

Ta có \(s_n\) hội tụ nên \(\lim\limits_{n\rightarrow+\infty}x_n=+\infty\)

Nếu \(2-\cos2\alpha\ne0\) thì

\(\lim\limits_{n\rightarrow+\infty}x_{n+1}=\lim\limits_{n\rightarrow+\infty}\dfrac{\left(2+\cos2\alpha\right)x_n+\cos^2\alpha}{\left(2-2\cos\alpha\right)x_n+2-\cos2\alpha}=\dfrac{2+\cos\alpha}{2-2\cos2\alpha}\), vô lí.

 Do đó \(2-2\cos2\alpha=0\) \(\Leftrightarrow\alpha=k\pi\left(k\inℤ\right)\)

 Với \(\alpha=k\pi\left(k\inℤ\right)\)

 Ta có \(x_{n+1}=3x_n+1\)  \(\Leftrightarrow2x_{n+1}+1=3\left(2x_n+1\right)=...=3^{n+1}\left(2x_1+1\right)=3^{n+1}\). Do đó:

 \(s_n=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^n}=\dfrac{1-\dfrac{1}{3^n}}{2}\)

Vậy nên \(\left(s_n\right)\) có giới hạn hữu hạn và \(\lim\limits_{n\rightarrow+\infty}=\dfrac{1}{2}\)