Another way to show convergence of $ \sum_{n=1}^{\infty} \frac{ (-1)^n }{n} $ We know the series
$$ \sum_{n=1}^{\infty} \frac{ (-1)^n }{n} $$
converges. The usual argument is to notice $b_n = \frac{1}{n}$ is decreasing, positive and converges to $0$, thus the series converges by the alternating test.
Q: Is there another way to show this series converges?
 A: 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 $|r|<1$, $\lim_{N\to\infty}r^N=0$.
A: 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!
A: 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.
A: I suggest to use the Dirichlet test. To conclude that the series
$$
\sum_{n=1}^{+\infty}a_nb_n
$$
converges, it is by this test sufficient to check that 


*

*the positive sequence $(a_n)$ is monotonically decreasing to $0$, which is clear for $a_n=1/n$, and 

*the sequence of partial sums $\sum_{n=1}^Nb_n$ is bounded. This is also clear in this case, since the partial sum is altering between $-1$ (if $N$ is odd) and $0$ (if $N$ is even).


The nice thing with this approach is that the Dirichlet test is quite strong, used sometimes in the theory of Fourier series, and easily proven with the summation by parts formula (which reminds of the integration by parts dito). It also opens up for an easy proof (but perhaps not that intuitive) of the special case that Leibniz test really is.
