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Math Help - Infinite series (3)

  1. #1
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    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}.
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  2. #2
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    Quote Originally Posted by NonCommAlg View Post
    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}
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    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!
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