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Prove given properties of infinite sums

by RoRi
  1. Given that

        \[ 1 + x^2 + x^4 + \cdots + x^{2n} + \cdots = \frac{1}{1-x^2} \qquad \text{if } |x| < 1, \]

    prove that

        \[ 1 + 0 + x^2 + 0 + x^4 + \cdots = \frac{1}{1-x^2} \qquad \text{if } |x| < 1. \]

  2. Use Theorem 10.2 (that \sum_{n=1}^{\infty} (\alpha a_n + \beta b_n) = \alpha \sum_{n=1}^{\infty} a_n + \beta \sum_{n=1}^{\infty} b_n), the result of part (a), and the fact that

        \[ 1 + x + x^2 + \cdots + x^n + \cdots = \frac{1}{1-x} \qquad \text{if } |x| < 1 \]

    to prove that

        \[ x + x^3 + x^5 + \cdots x^{2n+1} + \cdots = \frac{x}{1-x^2} \qquad \text{if } |x| < 1. \]

  3. Show that Theorem 10.2 when applied to

        \[ 1 + x + x^2 + \cdots + x^n + \cdots = \frac{1}{1-x} \qquad \text{if } |x| < 1 \]

    and

        \[ 1+x^2 + x^4 + \cdots + x^{2n} + \cdots = \frac{1}{1-x^2} \qquad \text{if } |x| < 1 \]

    does not yield

        \[ x + x^3 + x^5 + \cdots + x^{2n+1} + \cdots = \frac{x}{1-x^2} \qquad \text{if } |x| < 1. \]

    Instead it gives us the formula

        \[ \sum_{n=1}^{\infty} (x^n - x^{2n}) = \frac{1}{(1-x^2)} \qquad \text{for } |x| < 1. \]

  1. Proof. (Not sure if I’m missing something deeper here…) Since we are given that

        \[ 1 + x^2 + x^4 + \cdots + x^{2n} + \cdots = \frac{1}{1-x^2} \qquad \text{for } |x| < 1\]

    we have

        \[ 1 + 0 + x^2 + 0 + x^4 + 0 + \cdots + x^{2n} + 0 + \cdots = \frac{1}{1-x^2} \qquad \text{for } |x| < 1. \qquad \blacksquare \]

  2. Proof. Given that

        \[ 1 + x + x^2 + \cdots + x^n + \cdots = \frac{1}{1-x} \qquad \text{for } |x| < 1 \]

    we apply Theorem 10.2 to this and the result of part (a) so,

        \begin{align*} (1+x+x^2+\cdots + x^n + \cdots) - (1 + 0 + x^2 + 0 + x^4 + \cdots) &= \frac{1}{1-x} - \frac{1}{1-x^2}\\ \implies x + x^3 + x^5 + \cdots &= \frac{1-x^2-1+x}{(1-x)(1-x^2)} \\ &= \frac{-x(x-1)}{(1-x)(1-x^2)} \\ &= \frac{x}{1-x^2}. \qquad \blacksquare \end{align*}

  3. Proof. Applying Theorem 10.2 directly to the given expressions we have,

        \[ \sum_{n=0}^{\infty} x^n - \sum_{n=0}^{\infty} x^{2n} = \sum_{n=0}^{\infty} (x^n - x^{2n}) = \frac{x}{1-x^2}. \qquad \blacksquare \]

3 comments

  2. Mohammad Azad says:

    For (a) and (b) you should use partial sums,s_n = the sum from k=0 to k=n-1 of a_k where a_k=(x^n+(-x)^n)/2 is the nth partial sum of 1+0+x^2+0+… and s_n = (1/2)t_n + (1/2)d_n where t_n and d_n are the nth partial sums of (10.25) and (10.28) and so s_n -> (1/2)(1/(1-x)+1/(1+x))=1/(1-x^2)

  3. MathlessRick says:

    I guess the point of a) i is to show that their result is the same even though they are not identical (see exercise 24)
    But I cannot think of any solution but the obvious ones

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