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Formula for sum of the reciprocals of integers

  1. Give a definition of \sum_{k=m}^{m+n} a_k.
  2. Prove the following is true for n \geq 1:

        \[ \sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m}. \]


  1. We define

        \[ \sum_{k=m}^{m+n} a_k = a_m + a_{m+1} + \cdots + a_{m+n}.\]

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

        \[ \sum_{k=n+1}^{2n} \frac{1}{k} = \sum_{k=2}^2 \frac{1}{k} = \frac{1}{2}. \]

    On the right,

        \[ \sum_{m=1}^{2n} \frac{(-1)^{m+1}}{m} = \sum_{m=1}^2 \frac{(-1)^{m+1}}{m} = 1 - \frac{1}{2} = \frac{1}{2}. \]

    Thus, the formula holds for the case n = 1. Assume then that it holds for some n = j \in \mathbb{Z}_{\geq 1}. Then we have,

        \begin{align*}  &&\sum_{k=j+1}^{2j} \frac{1}{k} &= \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \\ \implies && \left(\sum_{k=j+1}^{2j} \frac{1}{k}\right)  + \frac{1}{2j+1} + \frac{1}{2j+2} &=  \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right)  + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \sum_{k=j+1}^{2(j+1)} \frac{1}{k} &= \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right)  + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \frac{1}{j+1} + \left(\sum_{k=j+2}^{2(j+1)} \frac{1}{k} \right)&= \left(\sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m}\right)  + \frac{1}{2j+1} + \frac{1}{2j+2} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \left( \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \right) + \frac{1}{2j+1} + \frac{1}{2(j+1)} - \frac{1}{j+1} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \left( \sum_{m=1}^{2j} \frac{(-1)^{m+1}}{m} \right) + \frac{1}{2j+1} - \frac{1}{2(j+1)} \\ \implies && \sum_{k=j+2}^{2(j+1)} \frac{1}{k} &= \sum_{m=1}^{2(j+1)} \frac{(-1)^{m+1}}{m}. \end{align*}

    Thus, the formula holds for j+1 if it holds for j; hence, we have shown that it holds for all n \in \mathbb{Z}_{\geq 1}. \qquad \blacksquare

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