# Infinite series (3)

• May 22nd 2009, 09:26 PM
NonCommAlg
Infinite series (3)
Evaluate $\displaystyle f(x)=\sum_{n=1}^{\infty} \frac{x^n}{(1+x)(1+x^2) \cdots (1+x^n)}, \ \ 1 \leq x \in \mathbb{R}.$
• May 23rd 2009, 01:42 AM
Isomorphism
Quote:

Originally Posted by NonCommAlg
Evaluate $\displaystyle f(x)=\sum_{n=1}^{\infty} \frac{x^n}{(1+x)(1+x^2) \cdots (1+x^n)}, \ \ 1 \leq x \in \mathbb{R}.$

$\displaystyle f(x)=\sum_{n=1}^{\infty} \frac{x^n}{(1+x)(1+x^2) \cdots (1+x^n)} = \sum_{n=1}^{\infty} u_{n-1} - u_n = u_0 - \lim_{n \to \infty} u_n = 1$ where $\displaystyle u_n = \frac{1}{(1+x)(1+x^2) \cdots (1+x^n)}$

$\displaystyle \lim_{n \to \infty} u_n = 0$ is obtained by squeezing, $\displaystyle 0 < \frac{1}{(1+x)(1+x^2) \cdots (1+x^n)} \leq \frac1{2^n}$
• May 23rd 2009, 02:30 AM
NonCommAlg
Quote:

Originally Posted by Isomorphism
$\displaystyle f(x)=\sum_{n=1}^{\infty} \frac{x^n}{(1+x)(1+x^2) \cdots (1+x^n)} = \sum_{n=1}^{\infty} u_{n-1} - u_n = u_0 - \lim_{n \to \infty} u_n = 1$ where $\displaystyle u_n = \frac{1}{(1+x)(1+x^2) \cdots (1+x^n)}$ and $\displaystyle \color{red}u_0=1.$

$\displaystyle \lim_{n \to \infty} u_n = 0$ is obtained by squeezing, $\displaystyle 0 < \frac{1}{(1+x)(1+x^2) \cdots (1+x^n)} \leq \frac1{2^n}$

correct! (Clapping)