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I am trying to compute a tight upper bound of the sum below.

$\sum_{i=1}^{n-1}n\frac{\frac{1}{3^i}}{\log_3{(n/3^i)}}$

I was able to 'simplify' it up to the expression below.

$n\sum_{i=1}^{n-1}\frac{1}{3^i(\log_3{n}- i)}$.

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    $\begingroup$ Why do you assume such a closed form exists? $\endgroup$
    – R. J. Mathar
    Commented Nov 9, 2023 at 8:56
  • $\begingroup$ I did not mean that. I only need to get a tight upper bound of the expression. $\endgroup$
    – ultrajohn
    Commented Nov 9, 2023 at 11:25
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    $\begingroup$ The terms start to become negative as soon as $i>\log_3(n)$. So an upper bound is given by the shorter sum where the upper limit $n-1$ is replaced by $\log_3(n)$. $\endgroup$
    – R. J. Mathar
    Commented Nov 10, 2023 at 9:24

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