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So I ran into this problem today. It asks me to use an identity to simplify the sum.

$$\sum_{j=7}^{27}\ln\left(\frac{j+1}{j}\right)$$

I have no idea where to start. I don't know any identity that fits this formation. Thanks.

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You can also look at the product; since $$A=\sum_{j=7}^{27}\ln\left(\frac{j+1}{j}\right)$$ then $$e^A=\prod_{j=7}^{27}\frac{j+1}{j}=\frac{8\times 9\times 10\times 11 \cdots \times 28}{7\times 8\times 9\times 10\times 11 \cdots \times 27}=\frac{(8\times 9\times 10\times 11 \cdots\times 27) \times 28}{7\times (8\times 9\times 10\times 11 \cdots \times 27)}$$ and notice how easily it simplifies. What is left is $$e^A=\frac{28}{7}=4$$ So, $A=2\log(2)$

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    $\begingroup$ thats imo more intuitive than identities. nice! $\endgroup$ Sep 13 '14 at 20:09
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Hint:

$$ \ln \dfrac a b = \ln a - \ln b$$

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  • $\begingroup$ Thanks... I simplified it to ln(28)-ln(7), is this the right answer? $\endgroup$
    – KFC
    Sep 13 '14 at 13:17
  • $\begingroup$ You can go one step further. $ \ln 28 = \ln (4 \times 7) = \ln 4 + \ln 7 $ $\endgroup$
    – Ishfaaq
    Sep 13 '14 at 13:19
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    $\begingroup$ I simplified it down to 2*ln(2) $\endgroup$
    – KFC
    Sep 13 '14 at 13:22
  • $\begingroup$ @user1763899: I think you are done. $\endgroup$
    – Ishfaaq
    Sep 13 '14 at 13:54
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Hint:

$$\ln\left(\frac{j+1}{j}\right)=\ln(j+1)-\ln(j)$$

This for $j>0$.

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