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Let $(x_n)$ be real sequences such that $x_{1}=\dfrac{1}{3}, x_{2n}=\dfrac{1}{3}x_{2n-1}, x_{2n+1}=\dfrac{1}{3}+x_{2n}, n=1,2,\cdots $.

Compute $$\lim_{x \to \infty} \sup x_{n} \text{ and } \lim_{x \to \infty} \inf x_{n}. $$

gt6989b
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Babymath
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2 Answers2

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Hint Notice that if $n$ is even, $$x_{n+2} = x_n/3 + 1/9$$ and if $n$ is odd $$x_{n+2} = x_n/3+1/3$$ with initial conditions $x_1 = 1/3, x_2 = 1/9$.

Analyze each of the two subsequences which are easy to do.

gt6989b
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    I proved :$\lim x_{2n}=\frac12$ and $\lim x_{2n+1}=\frac32$.But how to find $\sup x_n$ and $\inf x_n$ as $n\to +\infty$ – Babymath Mar 05 '14 at 13:09
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$$ x_{2n}=x_{2n-1}/3=(1/3+x_{2n-2})/3=x_{2n-2}/3+1/9 $$

$$ x{2n+1}=1/3+x_{2n}=1/3+x_{2n-1}/3 $$

and we have $x_1=1/3$, we can add $x_0=0$, which is ok by the definition

then sovle the recursive sequence, we get

$$ x_{2n} = \frac{1}{6}(1-\frac{1}{3^n}) $$

$$ x_{2n+1} = \frac{1}{2} - \frac{1}{6}\frac{1}{3^n} $$

so

$$ \lim_{n \to \infty} \sup x_{n} = 1/2 $$

$$ \lim_{n \to \infty} \inf x_{n} = 1/6 $$

delta
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