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Prove a formula for b^p – a^p and a resulting inequality

by RoRi
  1. Prove that for any p \in \mathbb{Z}_{>0} we have

        \[ b^p - a^p = (b-a)(b^{p-1} + b^{p-2}a + b^{p-3} a^2 + \cdots + ba^{p-2} + a^{p-1}). \]

  2. For p \in \mathbb{Z}_{>0} and n \in \mathbb{Z}_{>0} prove that

        \[ n^p < \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^p. \]

  3. Again for n,p \in \mathbb{Z}_{>0} prove

        \[ \sum_{k=1}^{n-1} k^p < \frac{n^{p+1}}{p+1} < \sum_{k=1}^n k^p. \]

  1. Proof. First we rewrite the sum on the right in summation notation, and then use the distributive property to compute,

        \begin{align*} (b-a) (b^{p-1} + b^{p-2}a + \cdots + &ba^{p-2} + a^{p-1}) \\ &= (b-a)\left(\sum_{k=0}^{p-1} b^{p-k-1} a^k \right) \\ &= b \left( \sum_{k=0}^{p-1} b^{p-k-1} a^k \right) - a \left( \sum_{k=0}^{p-1} b^{p-k-1}a^k \right) \\ &= \sum_{k=0}^{p-1} b^{p-k} a^k - \sum_{k=0}^{p-1} b^{p-k} a^{k+1} \\ &= \sum_{k=0}^{p-1} b^{p-k} a^k - \sum_{k=1}^p b^{p-k} a^k & (\text{reindexing 2nd sum})\\ &= b^p + \sum_{k=1}^{p-1} b^{p-k} a^k - \left(\left(\sum_{k=1}^{p-1} b^{p-k} a^k\right) + a^p \right) \\ &= b^p - a^p & (\text{sums cancel}). \end{align*}

    This is the identity requested. \qquad \blacksquare

  2. Proof. From the Binomial theorem we have

        \[ (n+1)^{p+1} = \sum_{k=0}^{p+1} \binom{p+1}{k} n^k. \]

    Thus,

        \begin{align*} \frac{(n+1)^{p+1} - n^{p+1}}{p+1} &= \left(\frac{1}{p+1}\right) \left( \left(\sum_{k=0}^{p+1} \binom{p+1}{k}n^k \right) - n^{p+1} \right)\\ &= \left( \frac{1}{p+1} \right) \left(n^{p+1} + (p+1)n^p + \left( \sum_{k=2}^{p+1} \binom{p+1}{k} n^k \right) - n^{p+1}\right)\\ &= n^p + \left(\frac{1}{p+1} \right) \left( \sum_{k=2}^{p+1} \binom{p+1}{k} n^k \right)\\ &> n^p \end{align*}

    since the second term is strictly positive for n,p positive integers. This establishes the inequality on the left.
    For the inequality on the right we use part (a) to get,

        \[ (n+1)^{p+1} - n^{p+1} = (n+1-n) \cdot ((n+1)^p + n(n+1)^{p-1} + \cdots + n^{p-1} (n+1) + n^p ). \]

    Thus,

        \begin{align*} \frac{(n+1)^{p+1} - n^{p+1}}{p+1} &= \left( \frac{1}{p+1} \right) \left((n+1)^p + n(n+1)^{p-1} + \cdots + n^{p-1}(n+1) + n^p\right) \\ &< \left( \frac{1}{p+1} \right) ((n+1)^{p+1} + (n+1)^p + \cdots + (n+1)^p) \\ &= \left( \frac{1}{p+1} \right) ((p+1)(n+1)^p) \\ & = (n+1)^p. \end{align*}

    In the second to last line we have just replaced each term n in the sum by n+1, giving the inequality. This establishes both sides of the requested inequality. \qquad \blacksquare

  3. Proof. The proof is by induction. For the case n= 1 we have,

        \[ \sum_{k=1}^{n-1} k^p = \sum_{k=1}^0 k^p =0, \qquad \frac{n^{p+1}}{p+1} = \frac{1}{p+1}, \qquad \sum_{k=1}^n k^p = \sum_{k=1}^1 k^p = 1. \]

    For a positive integer p, then 0 < \frac{1}{p+1} < 1; hence, the inequalities hold in the case n =1. Assume then that they hold for some n = m \in \mathbb{Z}_{>0}.
    For the left inequality,

        \begin{align*} \sum_{k=1}^{m-1} k^p < \frac{m^{p+1}}{p+1} &&\implies \sum_{k=1}^m k^p &< \frac{m^{p+1}}{p+1} + m^p \\ &&&< \frac{m^{p+1} + (m+1)^{p+1} - m^{p+1}}{p+1} &(\text{part (b)})\\ &&&= \frac{(m+1)^{p+1}}{p+1}. \end{align*}

    This establishes the left inequality for all n \in \mathbb{Z}_{>0}.
    For the right inequality, assume it is true for some n = m \in \mathbb{Z}_{>0}, then

        \begin{align*} \frac{m^{p+1}}{p+1} < \sum_{k=1}^m k^p \ &\implies& \frac{m^{p+1}}{p+1} + (m+1)^p &< \sum_{k=1}^{m+1} k^p \\ &\implies& \frac{m^{p+1} + (m+1)^{p+1} - m^{p+1}}{p+1} &< \sum_{k=1}^{m+1} k^p & (\text{part (b)})\\ &\implies& \frac{(m+1)^{p+1}}{p+1} &< \sum_{k=1}^{m+1} k^p. \end{align*}

    Hence, the right inequality is true for all n \in \mathbb{Z}_{>0}. Therefore, the inequalities requested are indeed true for all n \in \mathbb{Z}_{>0}. \qquad \blacksquare

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