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Show that the limit of ∑ 1/k = log (p/q) where the sum is from k = qn to pn

by RoRi
  1. For given integers p and q with 1 \leq q \leq p, prove

        \[ \lim_{n \to \infty} \sum_{k=qn}^{pn} \frac{1}{k} = \log \frac{p}{q}. \]

  2. Consider the series

        \[ 1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} - \frac{1}{8} + \ + \ + \ - \ - \ \cdots. \]

    This is a rearrangement of the alternating harmonic series (\sum \frac{(-1)^n}{n}) in which there are three positive terms followed by two negative terms. Prove that the series converges and that the sum is equal to \log 2 + \frac{1}{2} \log \frac{3}{2}.

Incomplete.

4 comments

  1. Evangelos says:

    Didn’t even realize there was a part (c) waiting for us on the next page before the start of section 10.23 lol. Here it goes.

    (c) Rearrange the alternating harmonic series, writing alternately p positive terms followed by q negative terms. Then, use part (a) to show that this rearranged series converges and has sum \log2 + \frac{1}{2} \log \frac{p}{q}.

    To show the above, we note that in the alternating harmonic series, every odd term is positive and every even term is negative. Now, consider the partial sum s_{m}, where m = (p+q)n for some integer n. This gives us the following:

        \begin{align*} s_{m} &= \sum_{k=1}^{pn} \frac{1}{2k-1} - \sum_{k=1}^{qn} \frac{1}{2k} \\ &= [\sum_{k=1}^{2pn} \frac{1}{k} - \sum_{k=1}^{pn} \frac{1}{2k}] - \sum_{k=1}^{qn} \frac{1}{2k} \end{align*}

    Using the results of parts (a) and (b), we get

        \begin{align*} [\sum_{k=1}^{2pn} \frac{1}{k} - \sum_{k=1}^{pn} \frac{1}{2k}] - \sum_{k=1}^{qn} \frac{1}{2k} &= \log 2 + \frac{1}{2}\log pn - \frac{1}{2}\log qn \\ &= \log 2 + \frac{1}{2} \log \frac{p}{q} \end{align*}

    • Evangelos says:

      The patterning of this problem is slightly different than that of the example in 10.21, but the principle is the same.

      (a) To show that

          \begin{align*} \lim_{n\to\infty} \sum_{k = qn}^{pn} \frac{1}{k} = \log\frac{p}{q} \end{align*}

      where p and q are integers, with

          \[1 \leq q \leq p\]

      , we first note that the above limit can be rewritten as follows

          \begin{align*} \lim_{n\to\infty} \sum_{k = qn}^{pn} \frac{1}{k} = \lim_{n\to\infty} \sum_{k = 1}^{pn} \frac{1}{k} - \sum_{k = 1}^{qn} \frac{1}{k} \end{align*}

      Then, using the limit relation from example 3, section 10.17

          \begin{align*} \lim_{n\to\infty} \sum_{k = 1}^{n} \frac{1}{n} = \log n + C + o(1) \end{align*}

      where C is Euler’s constant, we get the following

          \begin{align*} \lim_{n\to\infty} \sum_{k = 1}^{pn} \frac{1}{k} - \sum_{k = 1}^{qn} \frac{1}{k} &= [\log(pn) + C + o(1)] - [\log(qn) + C + o(1)] \\ &= \log\frac{p}{q} \end{align*}

      as requested.

      (b) To show that the sum of

          \begin{align*} 1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} - \frac{1}{8} +++-- ... \end{align*}

      approaches the value

          \[\log2 + \frac{1}{2}\log\frac{3}{2}\]

      as n goes to infinity, we first consider the partial sum,

          \[s_{5n}\]

      , which can be written as follows

          \begin{align*} s_{5n} &= \sum_{k = 1}^{3n} \frac{1}{2k - 1} - \sum_{k = 1}^{2n} \frac{1}{2k} \\ &= [\sum_{k = 1}^{6n} \frac{1}{k} - \sum_{k = 1}^{3n} \frac{1}{2k}] - \sum_{k = 1}^{2n} \frac{1}{2k} \end{align*}

      Using the results from part (a), as well as the limit relation from example 3, section 10.17, we get the following limit relation

          \begin{align*} [\sum_{k = 1}^{6n} \frac{1}{k} - \sum_{k = 1}^{3n} \frac{1}{2k}] - \sum_{k = 1}^{2n} \frac{1}{2k} &= \log 2 + \frac{1}{2} \log 3n - \frac{1}{2} \log 2n \\ &= \log 2 + \frac{1}{2} \log \frac{3}{2} \end{align*}

      • Evangelos says:

        Didn’t even realize there was a part (c) waiting for us on the next page before the start of section 10.23 lol. Here it goes.

        (c) Rearrange the alternating harmonic series, writing alternately p positive terms followed by q negative terms. Then, use part (a) to show that this rearranged series converges and has sum

            \begin{align*} \log2 + \frac{1}{2} \log \frac{p}{q} $. \end{align*}

        To show the above, we note that in the alternating harmonic series, every odd term is positive and every even term is negative. Now, consider the partial sum s_m, where m = (p+q)n for some integer n. This gives us the following:

            \begin{align*} s_{m} &= \sum_{k=1}^{pn} \frac{1}{2k-1} - \sum_{k=1}^{qn} \frac{1}{2k} \\ &= [\sum_{k=1}^{2pn} \frac{1}{k} - \sum_{k=1}^{pn} \frac{1}{2k}] - \sum_{k=1}^{qn} \frac{1}{2k} \end{align*}

        Using the results of parts (a) and (b), we get

            \begin{align*} [\sum_{k=1}^{2pn} \frac{1}{k} - \sum_{k=1}^{pn} \frac{1}{2k}] - \sum_{k=1}^{qn} \frac{1}{2k} &= \log 2 + \frac{1}{2}\log pn - \frac{1}{2}\log qn \\ &= \log 2 + \frac{1}{2} \log \frac{p}{q} \end{align*}

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