Another way to show convergence of $ \sum_{n=1}^{\infty} \frac{ (-1)^n }{n} $

One can collect the terms in pairs (since the terms go to $0$, this is okay) to get $$ \begin{align} \sum_{n=1}^\infty\left(\frac1{2n-1}-\frac1{2n}\right) &=\sum_{n=1}^\infty\frac1{2n(2n-1)}\\ &\le\frac12+\sum_{n=2}^\infty\frac1{2n(2n-2)}\\ &=\frac12+\frac14\sum_{n=2}^\infty\frac1{n(n-1)}\\ &=\frac12+\frac14\lim_{n\to\infty}\sum_{k=2}^n\left(\frac1{k-1}-\frac1k\right)\\ &=\frac12+\frac14\lim_{n\to\infty}\left(1-\frac1n\right)\\[6pt] &=\frac34 \end{align} $$ As user254665 comments, the terms of the series above are all positive, so the convergence is proven by the fact that the sum is bounded above by $\frac34$.

This is the negative of the series in the question, but a constant times a convergent series is convergent.

It's always safe to start with the partial sums, and to that, I will start with the geometric sum:

$$\frac{1-r^N}{1-r}=1+r+r^2+\dots+r^{N-1}$$

Since $r$ can be anything, we can treat it as a variable for integration:

$$\begin{align}\int_0^{-1}\frac{1-r^N}{1-r}dr&=\int_0^{-1}1+r+r^2+\dots+r^{N-1}dr\\&=\left.r+\frac12r^2+\frac13r^3+\dots+\frac1Nr^N\right|_0^{-1}\\&=-1+\frac12-\frac13+\dots+\frac{(-1)^N}N\end{align}$$

Note that the integral has no issues for $r\in(-1,0)$ and since $N\in\mathbb N$, we have no issues with complex numbers. Thus, the partial sums are given as

$$\sum_{n=1}^N\frac{(-1)^n}n=\int_0^{-1}\frac{1-r^N}{1-r}dr$$

Taking the limit to infinity, we have

$$\begin{align}\lim_{N\to\infty}\sum_{n=1}^N\frac{(-1)^n}n&=\lim_{N\to\infty}\int_0^{-1}\frac{1-r^N}{1-r}dr\\&=\int_0^{-1}\frac{1}{1-r}dr\\&=\left.-\ln(1-r)\right|_0^{-1}\\&=-\ln(2)\end{align}$$

where we use the fact that for $|r|<1$, $\lim_{N\to\infty}r^N=0$.

As shown by @robjohn, we can write the alternating harmonic series as

$$\sum_{n=1}^\infty\frac{(-1)^n}{n}=-\sum_{n=1}^\infty\left(\frac{1}{2n-1}-\frac1{2n}\right)$$

Next, we note that

$$\begin{align} \sum_{n=1}^{N}\left(\frac{1}{2n-1}-\frac{1}{2n}\right)&=\sum_{n=1}^{N}\left(\frac{1}{2n-1}+\frac{1}{2n}\right)-\sum_{n=1}^{N}\frac1n \tag 1\\\\ &=\sum_{n=1}^{2N}\frac1n -\sum_{n=1}^{N}\frac1n \tag 2\\\\ &=\sum_{n=N+1}^{2N}\frac1n \\\\ &=\sum_{n=1}^{N}\frac{1}{n+N} \\\\ &=\frac1N \sum_{n=1}^{N}\frac{1}{1+n/N} \tag 3 \end{align}$$

In going from $(1)$ to $(2)$ we simply noted that the sum, $\sum\limits_{n=1}^{2N}\frac1n$, can be written in terms of sums of even and odd indexed terms.

Finally, we observe that limit of $(3)$ is the Riemann sum for the integral $$\int_0^1 \frac{1}{1+x}\,dx=\log(2).$$

Putting it all together reveals

$$\sum_{n=1}^\infty\frac{(-1)^n}{n}=-\log(2)$$

And we are done!