# Using an identity to simplify the sum

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.

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, $A=2\log(2)$

• thats imo more intuitive than identities. nice! Sep 13 '14 at 20:09

Hint:

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

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

Hint:

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

This for $j>0$.