I have two questions!

I request you to give me detailed solutions for the two!!!

Thank you in advance!!!

Attachment 19155

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- Oct 2nd 2010, 01:56 AMlutesiumPlease help me solve these equations
I have two questions!

I request you to give me detailed solutions for the two!!!

Thank you in advance!!!

Attachment 19155 - Oct 2nd 2010, 02:00 AMEducated
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. - Oct 2nd 2010, 06:19 AMSoroban
Hello, lutesium!

Quote:

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

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

. . $\displaystyle \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) $

. . $\displaystyle =\;\ln(n+1)$

- Oct 2nd 2010, 09:09 AMProve It
$\displaystyle e^{-x} + 1 = 0$

$\displaystyle e^{-x} = -1$

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

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

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

$\displaystyle -x = 0 + i\pi$

$\displaystyle -x = i\pi$

$\displaystyle x = -i\pi$. - Oct 2nd 2010, 09:20 AMlutesiumThanks
Thank you very much for your detailed answers!!!

Thanks again!!!