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Prove inequalities of the logarithm with respect to some series

by RoRi

Consider a partition P = \{ a_0, a_1, a_2, \ldots, a_n \} of the interval [1,x] for some x > 1.

  1. Find step functions that are constant on the open subintervals of P and integrate to derive the inequalities:

        \[ \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) < \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right). \]

  2. Give a geometric interpretation of the inequalities in part (a).
  3. Find a particular partition P (i.e., choose particular values for a_0, a_1, \ldots, a_n) to establish the following inequalities for n > 1,

        \[ \sum_{k=2}^n \frac{1}{k} < \log n < \sum_{k=1}^{n-1} \frac{1}{k}. \]

  1. Proof. We define step function s and t by

        \begin{align*} s(x) &= \frac{1}{a_k} & \text{for } x &\in [a_{k-1}, a_k ) \\ t(x) &= \frac{1}{a_{k-1}} & \text{for } x &\in [a_{k-1}, a_k ). \end{align*}

    Since \frac{1}{x} is strictly decreasing on \mathbb{R}_{>0}, we have

        \[ s(x) < \frac{1}{x} < t(x) \qquad \text{for all } x \in \mathbb{R}_{>0}. \]

    Therefore, using the definition of the integral of a step function as a sum,

        \begin{align*} && \int_1^x s(u) \, du &< \int_1^x \frac{1}{u} \, du < \int_1^x t(u) \, du \\ \implies && \sum_{k=1} s_k (a_k - a_{k-1} ) &< \log x < \sum_{k=1}^n t_k (a_k - a_{k-1}) \\ \implies && \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) &< \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right). \qquad \blacksquare \end{align*}

  2. Geometrically, these inequalities say that the area under the curve \frac{1}{x} lies between the step functions that take on the values \frac{1}{a_k} and \frac{1}{a_{k-1}} for each x \in [ a_{k-1}, a_k ).
  3. Proof. To establish these inequalities we pick the partition,

        \[ P = \{ 1, 2, 3, \ldots, n \} \qquad \text{for } n > 1. \]

    Then, applying part (a) we have

        \begin{align*} && \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) &< \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right) \\ \implies && \sum_{k=1}^n \frac{1}{a_k} &< \log n < \sum_{k=1}^n \frac{1}{a_{k-1}} &(\text{since } a_k - a_{k-1} = 1) \\ \implies && \sum_{k=2}^n \frac{1}{k} &< \log n < \sum_{k=1}^{n-1} \frac{1}{k}. \end{align*}

    The final line follows since a_0 = 1, a_1 = 2, \ldots so the sum on the left starts with \frac{1}{2} and the sum on the right only runs to \frac{1}{n-1}. These were the inequalities requested. \qquad \blacksquare

2 comments

  1. Anonymous says:

    In part c, if you define P = {1,2,3, …, n} then only a_0 through a_(n-1) are defined, the left sum refers to a_n which is not defined in P.

    • William says:

      For the summation symbol in the first two lines, I think it should be n-1 instead of n (since there are n-1 terms. Shifting to k=2 on the left side would turn the n-1 into an n

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