# Math Help - Infinite series (3)

1. ## Infinite series (3)

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

2. Originally Posted by NonCommAlg
Evaluate $f(x)=\sum_{n=1}^{\infty} \frac{x^n}{(1+x)(1+x^2) \cdots (1+x^n)}, \ \ 1 \leq x \in \mathbb{R}.$
$f(x)=\sum_{n=1}^{\infty} \frac{x^n}{(1+x)(1+x^2) \cdots (1+x^n)} = \sum_{n=1}^{\infty} u_{n-1} - u_n = u_0 - \lim_{n \to \infty} u_n = 1$ where $u_n = \frac{1}{(1+x)(1+x^2) \cdots (1+x^n)}$

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

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

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