Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
how did you get at this result? – user 1618033 Jul 10 '12 at 19:10

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 $\lim_{x \rightarrow \infty} g(x) - x = \lim_{x \rightarrow \infty} \frac{x}{x-1} = 1$

share|cite|improve this answer
the limit may be easily found by applying AM-GM, as well. – user 1618033 Jul 11 '12 at 9:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.