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While solving a problem based on integration, I arrived at the following

$$\sum\limits_{x = 1}^{38} \ln\left(\frac{x}{x+1}\right)$$

I'm supposed to prove that this is less than $\ln(99)$ in order to do that, I tried evaluating the summation on wolfram(as I don't know how to compute this by myself) , but in the general form, wolfram uses the gamma function, which I don't know.

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    $\begingroup$ Hint: en.wikipedia.org/wiki/Telescope :) $\endgroup$
    – Martin R
    Commented Sep 25, 2022 at 17:21
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    $\begingroup$ Or look here: math.stackexchange.com/a/1747383/42969 $\endgroup$
    – Martin R
    Commented Sep 25, 2022 at 17:23
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    $\begingroup$ $\ln\left(\frac{x}{x+1}\right) = \ln(x) - \ln(x+1)$. Do you know what a “telescoping sum“ is? $\endgroup$
    – Martin R
    Commented Sep 25, 2022 at 17:26
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    $\begingroup$ Btw, all terms in the sum are negative, therefore the sum is surely less than $\ln(99)$. $\endgroup$
    – Martin R
    Commented Sep 25, 2022 at 17:27
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    $\begingroup$ @MartinR, regarding you fist comment, informally, I go, but the only kind I'' familiar with is an infinite GP $\endgroup$
    – math and physics forever
    Commented Sep 25, 2022 at 17:29

1 Answer 1

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This is a telescoping series: $$ \sum_{x=1}^{38}\ln\frac{x}{x+1} = \sum_{x=1}^{38}\ln x -\ln(x+1) \\ = (\ln 1-\ln 2)+(\ln 2-\ln 3)+\cdots+(\ln 38-\ln 39) \\ = \ln 1 - \ln 39 \\ = -\ln 39. $$

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  • $\begingroup$ oh,OK. Thanks!! $\endgroup$
    – math and physics forever
    Commented Sep 25, 2022 at 18:18
  • $\begingroup$ @fhhh : If this answers your question, please select the checkmark next to the question to accept the answer. You can +1 if you particularly like the answer, but that is not necessary. $\endgroup$
    – Disintegrating By Parts
    Commented Sep 27, 2022 at 17:14
  • $\begingroup$ I already upvoted, but I was waiting for a day just incase, thanks and sorry $\endgroup$
    – math and physics forever
    Commented Sep 27, 2022 at 17:17

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