Taylor Series/Sum problem 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!
 A: Hints: for $\;|x|<1\;$
$$\frac1{x(1+x)}=\sum_{n=0}^\infty(-1)^nx^{n-1}\implies \log\frac x {1+x}=\sum_{n=0}^\infty\frac{(-1)^n x^n}n\;\;\ldots$$
A: 
I thought it might be instructive to present an approach that relies on only straightforward arithmetic and the identification of a Riemann sum. To that end, we now proceed.

Note that we can write
$$\begin{align}\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n}&=\sum_{n=1}^N \left( \frac{1}{2n-1}-\frac{1}{2n}\right)\\\\
&=\sum_{n=1}^N\left( \frac{1}{2n-1}+\frac{1}{2n}\right)-2\sum_{n=1}^N\frac{1}{2n}\\\\
&=\sum_{n=1}^{2N}\frac1n -\sum_{n=1}^N\frac1n\\\\
&=\sum_{n=N+1}^{2N}\frac1n\\\\
&=\sum_{n=1}^N\frac{1}{N+n}\\\\
&=\frac1N \sum_{n=1}^N\frac1{1+n/N}\tag 1
\end{align}$$
Recognizing that the limit of $(1)$ is the Riemann sum for $\int_0^1 \frac1{1+x}\,dx=\log(2)$, we find the limit of the alternating harmonic series is $\log(2)$.

Tools used: Arithmetic and Riemann Sums

A: 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,
$$=-\int_{0}^{1} \frac{1}{1+x} dx$$
$$=-\ln 2$$
