• October 2nd 2010, 01:56 AM
lutesium
I have two questions!
I request you to give me detailed solutions for the two!!!

Thank you in advance!!!

Attachment 19155
• October 2nd 2010, 02:00 AM
Educated
Why do you want us to give you detailed solutions to them? Have you attempted them yourself? Can you show us some of your working?

If you want to learn how to solve them or you are stuck at solving them, that's fine, we'll help you. We'll help you in understanding it or get past whatever you are stuck on.
But we cannot just give you full detailed solutions for whatever other reason you might have, and that includes homework.
• October 2nd 2010, 06:19 AM
Soroban
Hello, lutesium!

Quote:

$\displaystyle \text{Evaluate: }\;S \;=\;\sum^n_{k=1}\ln\left(1 + \frac{1}{k}\right)$

$\displaystyle S \;=\;\ln\left(1 + \frac{1}{1}\right) + \ln\left(1 + \frac{1}{2}\right) + \ln\left(1 + \frac{1}{3}\right) + \ln\left(1 + \frac{1}{4}\right) + \hdots + \ln\left(1 + \frac{1}{n}\right)$

. . $\displaystyle =\;\ln(2) + \ln\left(\frac{3}{2}\right) + \ln\left(\frac{4}{3}\right) + \ln\left(\frac{5}{4}\right) + \hdots + \ln\left(\frac{n+1}{n}\right)$

. . $\displaystyle =\;\ln\left(2\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\:\cdots\;\frac{n+1}{n}\right)$

. . $\displaystyle =\;\ln\left(\frac{\rlap{/}2}{1}\cdot\frac{\rlap{/}3}{\rlap{/}2}\cdot\frac{\rlap{/}4}{\rlap{/}3}\cdot\frac{\rlap{/}5}{\rlap{/}4}\:\cdots\;\frac{n+1}{\rlap{/}n}\right)$

. . $=\;\ln(n+1)$
• October 2nd 2010, 09:09 AM
Prove It
$e^{-x} + 1 = 0$

$e^{-x} = -1$

$-x = \log{(-1)}$

$-x = \ln{|-1|} + i\arg{(-1)}$

$-x = \ln{1} + i\pi$

$-x = 0 + i\pi$

$-x = i\pi$

$x = -i\pi$.
• October 2nd 2010, 09:20 AM
lutesium
Thanks
Thank you very much for your detailed answers!!!

Thanks again!!!