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Let $(a_n)_{n \geq 0}$ be a sequence with $a_0,a_1 \in (0,1)$ and $$a_{n+1}=a_n^2a_{n-1}-a_na_{n-1}+1$$ Find $$\lim_{n \to \infty} (a_0 \cdot a_1 \cdot ... \cdot a_n)$$

I managed to prove that $a_n \in (0,1)$ and that the sequence is increasing and convergent to $1$. Then I denoted $b_n=a_0 \cdot a_1 \cdot ... \cdot a_n$ and since it is strictly decreasing and bounded, it must be convergent to a number $l$.

Using the relation given I obtained $$b_{n+1}=\frac{b_n^3}{b_{n-1}b_{n-2}}-\frac{b_n^2}{b_{n-2}}+b_n \iff b_{n+1}-b_n=\frac{b_n^2}{b_{n-1}b_{n-2}}(b_n-b_{n-1})$$ and thus got $$b_{n+1}-b_n=\frac{b_n^2 \cdot b_{n-1}}{b_1\cdot b_0}(b_2-b_1)$$ Applying limits yields $0=\frac{l^3(b_2-b_1)}{b_1b_0}$, hence $l=0$.

My question is whether this problem could have been done differently, maybe without the notation of $b_n$. At first I tried to write $$a_0\cdot a_1 \cdot ... \cdot a_n <a_n^{n+1}$$ and then $\lim_{n \to \infty}a_n^{n+1}=e^{\lim_{n \to \infty} (a_n-1)(n+1)}$ and tried to prove that $$\lim_{n \to \infty}n(a_n-1)=-\infty$$ using Cesaro-Stolz, but didn't get anything...

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    $\begingroup$ @mathlover $\lim_{n\to\infty}\left(1-\frac 1n\right)^n=1/e\ne 0$. So, your statement is false. $\endgroup$ – Mark Viola Nov 20 '17 at 15:36
  • $\begingroup$ @MathLover Not necessarily. For instance, the product of all possible $\frac{n^2}{n^2+1}$ for $n$ from $1$ and up gives a result greater than $\frac14$ in the end. $\endgroup$ – Arthur Nov 20 '17 at 15:37
  • $\begingroup$ If the infinite sequence of product has a maximum value less than $1$ @MathLover statement is true $\endgroup$ – avz2611 Nov 20 '17 at 15:38
  • $\begingroup$ I think is obvious $\lim_{n \to \infty}n(a_n-1)=-\infty$ since you proved that $\lim_{n \to \infty}a_n=1$ and $0<a_n<1$ $\endgroup$ – Guy Fsone Nov 20 '17 at 15:58
  • $\begingroup$ @avz2611 That's not relevant here. $\endgroup$ – Mark Viola Nov 20 '17 at 15:58
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$$a_{n+1}=a_n^2a_{n-1}-a_na_{n-1}+1\Longleftrightarrow\frac{1-a_{n+1}}{1-a_n}=a_na_{n-1}$$

By telescopic product we have, $$\frac{a_{n+1}-1}{a_1-1}= \prod_{k=1}^{n}\frac{1-a_{k+1}}{1-a_k}=\prod_{k=1}^{n}(a_ka_{k-1}) = \prod_{k=1}^{n}a_k\prod_{k=1}^{n}a_{k-1} = a_0\prod_{k=0}^{n}a_k\prod_{k=0}^{n-1}a_{k} $$

Taking limit on both side give,

$$\lim_{n\to \infty} (1-a_n) = a_0(1-a_1)\left(\prod_{k=0}^{\infty}a_k\right)^2$$ Thus, $$\prod_{k=0}^{\infty}a_k = \sqrt{\frac{\lim\limits_{n\to \infty} (1-a_n)}{a_0(1-a_1)}}=0$$ since you proved that $$\lim\limits_{n\to \infty} a_n=1$$

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