# Thread: Infinite series (3)

1. ## Infinite series (3)

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

2. 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, $\displaystyle 0 < \frac{1}{(1+x)(1+x^2) \cdots (1+x^n)} \leq \frac1{2^n}$

3. 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!