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.
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
1
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, $A=2\log(2)$
answered Sep 13, 2014 at 14:10
-
2$\begingroup$ thats imo more intuitive than identities. nice! $\endgroup$ Sep 13, 2014 at 20:09
$\begingroup$
$\endgroup$
4
Hint:
$$ \ln \dfrac a b = \ln a - \ln b$$
$\begingroup$
$\endgroup$
Hint:
$$\ln\left(\frac{j+1}{j}\right)=\ln(j+1)-\ln(j)$$
This for $j>0$.
