Question: Find the value of $\displaystyle\sum\_{n=1}^\infty \frac{(-1)^n}{n}$.

I have no idea how to start. I've tried listing out a few terms and I still can't find a pattern or anything. Any whatsoever help is appreciated!

Answer: 
Hints: for $\;|x|<1\;$

$$\frac1{x(1+x)}=\sum\_{n=0}^\infty(-1)^nx^{n-1}\implies \log\frac x {1+x}=\sum\_{n=0}^\infty\frac{(-1)^n x^n}n\;\;\ldots$$
Note that,

$$\int\_{0}^{1} x^{n-1} dx=\frac{1}{n}$$

So that,

$$\sum\_{n=1}^{\infty} \frac{(-1)^n}{n}$$

$$=\sum\_{n=1}^{\infty} \left( \int\_{0}^{1} x^{n-1} dx \right)(-1)^n$$

Interchange the sum and integral.

$$=\int\_{0}^{1} \sum\_{n=1}^{\infty} x^{n-1}(-1)^{n-1} (-1) dx$$

$$=-\int\_{0}^{1} \sum\_{n=1}^{\infty} x^{n-1}(-1)^{n-1} dx$$

$$=-\int\_{0}^{1} \sum\_{n=0}^{\infty} (-x)^n dx$$

Recognize the geometric series to get,

$$=-\int\_{0}^{1} \frac{1}{1+x} dx$$

$$=-\ln 2$$