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I want to calculate the result of $$\sum\limits_{k=1}^{\log_{2}{n}}\log_{3}{\frac{n}{2^k}}$$ I used two below approaches. Both approaches are based on $\log A + \log B = \log (A \times B)$ and $\sum\limits_1^n i = \frac{n(n+1)}{2}$ rules:

  • $$ \begin{align} \sum\limits_{k=1}^{\log_{2}{n}}\log_{3}{\frac{n}{2^k}} &= \log_{3}{\frac{n^{\log_{2}n}}{2^{1+2+...+\log_{2}{n}}}} \tag1\\ &= \log_{3}{\frac{n^{\log_{2}n}}{2^{(\log_{2}n) (\log_{2}{n}\ +\ 1)}}} \tag2\\ &= \log_{3}{(\frac{n}{2^{(\log_{2}{n})\ +\ 1}})^{{\log_{2}n}}} \tag3\\ &= ({\log_{2}n})\times \log_{3}{(\frac{n}{2^{(\log_{2}{n})\ +\ 1}})} \tag4 \\ &= ({\log_{2}n})\times \log_{3}{(\frac{n}{2n})} \tag5\\ &= ({\log_{2}n})\times \log_{3}{(\frac{1}{2})} \tag6 \end{align}$$

  • In this approach I use Change of variables: Get $m=\log_2 n$. So $$\begin{align} \sum\limits_{k=1}^{\log_{2}{n}}\log_{3}{\frac{n}{2^k}} &= \sum\limits_{k=1}^{m}\log_{3}{\frac{2^m}{2^k}} \tag7 \\ &= \sum\limits_{k=1}^{m}\log_{3}{2^{m-k}} \tag8 \\ &= \sum\limits_{k=1}^{m}(m-k)\log_{3}{2} \tag9 \\ &= (m^2 - \frac{m(m+1)}{2})\log_{3}{2} \tag{10} \\ &= (\frac{m^2 - m}{2})\log_{3}{2} \tag{11} \end{align}$$ By substituting $\log_2 n$ instead of $m$, we will have: $$ \sum\limits_{k=1}^{\log_{2}{n}}\log_{3}{\frac{n}{2^k}} = (\frac{{(\log_2 n)}^2 - {\log_2 n}}{2})\log_{3}{2} \tag{12}$$

As you can see the result of first approach is negative, so it must be wrong and the second approach must be true; but what is wrong in the first approach?

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  • $\begingroup$ The summation, (1), evaluates to $\frac{(\log_{2}(n) - 1) \, \log(n)}{2 \, \log(3)}. $ $\endgroup$
    – Leucippus
    Nov 20, 2022 at 10:33
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    $\begingroup$ A way to find your mistake yourself: Substitute a value (say $n=4$) in each of the steps (1) to (6) to find where the error is. $\endgroup$
    – GEdgar
    Nov 20, 2022 at 11:28
  • $\begingroup$ The sum $1+2+\cdots+\log_{2}(n)=\dfrac{\log_{2}(n)\left(\log_{2}(n)+1\right)}{2}$ $\endgroup$ Nov 28, 2022 at 17:16
  • $\begingroup$ Thank you very much for your time. That was a mistake caused by carelessly. $\endgroup$ Dec 5, 2022 at 7:53

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