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: 
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> 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.
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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 $\log(2)$.

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> Tools used: Arithmetic and Riemann Sums
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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,

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

$$=-\ln 2$$