How to prove that the series $\sum_{n=1}^\infty \frac{(-1)^n}n$ converges I know by definition that a series $$\sum_{n=1}^\infty a_n$$ converges when the sequence of partial sums $S_N=a_1 + a_2 + .. + a_N$ converges to $S$, so $\lim_{N\rightarrow \infty} S_N=S$. 
So in particular, I'm given the series $$\sum_{n=1}^\infty \frac{(-1)^n}{n},$$ which converges to $\ln 2$.
I'm kinda stuck on how to get started. So far I have,
$\forall \epsilon >0$, $\exists N>0$ s.t. if $n>N$ then $|a_n-0|<\epsilon$.
Is this ok so far? How do I go from here?
 A: Your series does not converge to zero - the sequence it sums does, though. (You don't need to prove this, but your series sums to $-\text{ln}(2)$.)
Hint: Try grouping successive terms together. In particular, this becomes
$$\sum_{n=0}^\infty \left(-\frac 1 {2n+1} + \frac 1 {2n+2}\right)$$
See if you can work with that.
A: Now that you know the answer is $\text{ln}(2)$, perhaps you want to try some kind of estimate using Taylor's Theorem... Note that $\xi>0$ and so $1+\xi > 1$. $$|\sum_{k=1}^{n} \frac{(-x)^k}{k}-\text{ln}(1+x)| \leq \frac{\frac{n!}{(1+\xi)^n}\cdot x^{n+1}}{(n+1)!} < \frac{x^{n+1}}{n+1}$$
So in our case (it works in all possible values of $x$, since $x$ itself is fixed), we have $x=1$ and hence we have an upper bound of our error of $\frac{1}{n+1}$. Hence, as $n \rightarrow \infty$, $$\sum_{k=1}^{n} \frac{(-x)^k}{k}-\text{ln}(1+x) \rightarrow 0.$$
As for showing the sequence, $[S_{1},S_{2},...$, is convergent, just look at how $S_{1} < S_{3} < ... S_{2n+1}$ and $S_{2} > S_{4} > ... S_{2n}$ and combine to show that $S_{1} < S_{3} < ... S_{2n+1} < S_{2n} < S_{2n-2} < ... S_{4} < S_{2}$. Now, $\forall k,l>2n+1$, $|S_{k}-S_{l}| < S_{2n} - S_{2n+1} = \frac{1}{2n}$. So choose $n>\frac{1}{2\varepsilon}$ to complete the proof.
AFTERTHOUGHT: A subtle point is that I've proven Cauchy and not convergence here; however, in $\mathbb{R}$ they are equivalent so the proof still holds.
A: The following lemma (known in some circles as Leibniz's test) trivially solves the problem.

Let $(a_n)_{n=1}^{\infty}$ be a monotonic sequence of real numbers converging to $0$. Then the following series
  $$ \sum_{n=1}^{\infty} (-1)^n a_n $$
  converges.

I leave out the proof, but it can be found on Wikipedia. 
