Show $\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}\,\forall x\in(-1,1)$. Show $\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}\,\forall x\in(-1,1)$. Which value does $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ take?

Now because I skipped forward in my (personal) textbook I know that I could tackle this using knowledge of the Maclaurin/Taylor series. However, it was not covered in the lecture (yet) and my mind is fixated on using Maclaurin/Taylor (which I'm not allowed to use)! Can anybody show me an alternate approach that I will probably feel very stupid for not seeing?
 A: Hint: A term by term integration to the Geometric Series $\frac{1}{1-x}=\sum\nolimits_{k=0}^{\infty }{{{x}^{k}}}$ 
A: You have the series $\sum_{k=1}^{\infty}\frac{1}{k}x^k$ which is a power series. The radius of convergence is calculated by $\frac{1}{R} = \limsup\limits_{k\rightarrow \infty} |\frac{1}{k^{1/k}}| = 1$ and thus the series converges absolutely $\forall x\in (-1,1)$. Therefore we can differentiate the series termwise. I.e.
$\frac{d}{dx}\sum_{k=1}^{\infty}\frac{1}{k}x^k = \sum_{k=1}^{\infty}x^{k-1} = \frac{1}{1-x}$.
Thus,
$\sum_{k=1}^{\infty}\frac{1}{k}x^k = -ln(1-x) \forall x \in (-1,1)$.
A: Use the geometric series:
$$\sum_{n=0}^{+\infty} x^n =\frac{1}{1-x} \qquad \text{ for } x\in (-1,1)$$
We can integrate both sides:
$$\int\sum_{n=0}^{+\infty} x^n\,dx=-\ln(1-x)$$
Because the geometric series converges uniformly on $(-1,1)$, we can exchange integral and summation:
\begin{align}
-\ln(1-x)&=\sum_{n=0}^{+\infty} \int x^n\,dx \\
&=\sum_{n=0}^{+\infty} \frac{x^{n+1}}{n+1} \\
&=\sum_{n=1}^{\infty} \frac{x^n}{n}
\end{align}
For your second questions, we can send $x \mapsto -x$ to find that 
$$-\ln(1+x)=\sum_{n=1}^{+\infty} \frac{(-1)^n x^n}{n}$$
Letting $x\ \to 1^{-}$,  
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n}=-\ln 2$$
