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Let $\{x_{n}\}_{n=1}^\infty$, with $x_{1}=a$ where $a>1$ be a sequence that satisfies the relation:

$$ x_{1}+x_{2}+...+x_{n+1}= x_{1}x_{2}\cdots x_{n+1}$$

For this problem, the requirement is to prove that $x_{n}$ is convergent, and then find its limit when $n$ goes to $\infty$. I think I can handle with these two requirements, but my curiosity is related to the way $x_{n}$ looks like and wonder if there is a nice closed form to it.

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up vote 4 down vote accepted

For $n \ge 2$, $x_n = \dfrac{a^{2^{n-2}}}{P_n(a)}$ where $\displaystyle P_{n}(a) = a^{2^{n-2}} - \prod_{j=2}^{n-1} P_j(a)$ is a polynomial in $a$ of degree $2^{n-2}$.

$$\eqalign{P_2(a) &= a-1 \cr P_3(a) &= a^2-a+1 \cr P_4(a) &= a^4-a^3+2 a^2-2 a+1\cr P_5(a) &= a^8-a^7+3a^6-6a^5+9a^4-10a^3+8a^2-4a+1\cr}$$

It looks like:

The coefficient of $a^0$ in $P_n(a)$ is $1$ for $n \ge 3$.

The coefficient of $a^1$ in $P_n(a)$ is $-2^{n-3}$ for $n \ge 3$.

The coefficient of $a^2$ in $P_n(a)$ is $2^{2n-7}$ for $n \ge 4$.

The coefficient of $a^3$ in $P_n(a)$ is $\dfrac{2^{n-3}-8^{n-3}}{6}$ for $n \ge 3$.

EDIT: The coefficient of $a^4$ in $P_n(a)$ is $\dfrac{2^n}{32} - \dfrac{4^n}{384} + \dfrac{16^n}{98304}$ for $n \ge 5$.

All this should be provable by induction. I got these by looking at the first few members of the sequence using Maple.

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how did you get at this result? –  Chris's sis Jul 10 '12 at 19:10
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The sum $s_n := x_1 + ... + x_n$ satisfies $s_n = g(s_{n-1})$ where $g(x) = \frac{x^2}{x-1}$ (prove this by induction). Since $g(x) > x + 1$ for all $x > 1$, the sequence $s_n$ tends to infinity. So you have $$x_n = g(s_{n-1}) - s_{n-1},$$ which tends to $\lim_{x \rightarrow \infty} g(x) - x = \lim_{x \rightarrow \infty} \frac{x}{x-1} = 1$

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the limit may be easily found by applying AM-GM, as well. –  Chris's sis Jul 11 '12 at 9:02
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