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$\sum_{k=2}^n \log(\frac{k}{k+1})$

So far I have: $$\sum_{k=2}^n\log(k) -\log(k+1) \\ \text{Looking at this I can see that this may be a telescoping series.} \\ \sum_{k=2}^n\log(k) -\log(k+1) = [\log(2) - \log(3)] +[\log(3) - \log(4)] + \cdots + [\log(n) - \log(n+1)] \\ \text{What is in red cancels.}\\ [\log(2) - \color{red}{\log(3)}] +[\color{red}{\log(3)} - \color{red}{\log(4)}] + \cdots + [\color{red}{\log(n)} - \log(n+1)] \\ \therefore \sum_{k=2}^n \log(\frac{k}{k+1}) = \log(2) - \log(n+1)?$$

I don't know if this is correct. I tried to check it with summation calculators and they say it diverges (which I do see why), but is this how you should go about evaluating the sum?

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    $\begingroup$ It is how you should go about it, but you have a small mistake. Look carefully at the last term. $\endgroup$
    – Daniel Fischer
    Sep 26, 2020 at 21:08
  • $\begingroup$ You are almost right and the summation calculators are also right, but in different ways. Your question is about a partial sum. These calculators calculate the infinite sum (which from your result it will diverge to $-\infty$) $\endgroup$
    – player3236
    Sep 26, 2020 at 21:08
  • $\begingroup$ It should be $log(n+1)$ I am assuming. What that the mistake? Always good to double check even though it may be silly. $\endgroup$
    – user750949
    Sep 26, 2020 at 21:16
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    $\begingroup$ Yes, that was the mistake. $\endgroup$
    – Daniel Fischer
    Sep 26, 2020 at 21:21

1 Answer 1

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Let $$ S = \sum_{k=2}^n \ln \left( \frac{k}{k+1} \right) \text{.} $$ Then \begin{align*} \mathrm{e}^S &= \exp\left( \sum_{k=2}^n \ln \left( \frac{k}{k+1} \right) \right) \\ &= \prod_{k=2}^n \exp \left( \ln \left( \frac{k}{k+1} \right) \right) & [ \mathrm{e}^{A+B} = \mathrm{e}^A \cdot \mathrm{e}^B] \\ &= \prod_{k=2}^n \frac{k}{k+1} \end{align*} and you should have no trouble finishing this telescoping product. What is the denominator in the last multiplicand?

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    $\begingroup$ Why not obtain the telescoping product directly with the log properties? $\endgroup$
    – Bernard
    Sep 26, 2020 at 21:21
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    $\begingroup$ @Bernard : Amount of notation obscuring the cancellation. Also, somewhere along the way, "sum of logs" ${} \implies {}$ "exponentiate" became a first reaction. $\endgroup$
    – Eric Towers
    Sep 26, 2020 at 21:24

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