Infinite series (3)

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May 22nd 2009, 09:26 PM
NonCommAlg

Infinite series (3)

Evaluate $f(x)=\sum_{n=1}^{\infty} \frac{x^n}{(1+x)(1+x^2) \cdots (1+x^n)}, \ \ 1 \leq x \in \mathbb{R}.$
May 23rd 2009, 01:42 AM
Isomorphism

Quote:

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, $0 < \frac{1}{(1+x)(1+x^2) \cdots (1+x^n)} \leq \frac1{2^n}$
May 23rd 2009, 02:30 AM
NonCommAlg

Quote:

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

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