Question about the convergence of $\sum_{n=1}^\infty (-1)^n$ This series obviously converges because $\sum_{n=1}^\infty (-1)^n = -1+1-1+1-1+1-1...=0$, but if the series $\sum_{n=1}^\infty a_n$ converges $\implies\lim{a_n}=0$, but in this case, $a_n=(-1)^n$ doesn't converge to 0 because $\lim{a_{2n}}=1$ and $\lim{a_{2n-1}}=-1\implies\lim{a_n}\nexists$
What am I missing here?
 A: 
This series obviously converges

No it doesn't. The definition of convergence is that it approaches some value. In this case it keeps alternating between 0 and -1; it is not approaching anything.
A: You are wrong from the start. Why do you assert that$$-1+1-1+1-1+\cdots=0?$$ If you are adding a finite number of terms, then you will get $0$ half of the times and $-1$ the other half.
A: This is Grandi's series, but with $-1$ added to the beginning. Its value depends on the axioms you are using.
If you consider the partial sums, they will alternate between $-1$ and $0$, and thus the series is divergent and has no value.
Alternatively, we can use Cesàro summation, which requires finding $$\lim_{k\to\infty}(\frac{\sum_{n=1}^k (-1)^n}{k})$$
This is the average of the partial sums, which will evaluate to $-1/2$ in this case. We can also reach this value by considering that the series is a geometric progression with $a = -1$, $r = -1$ and therefore infinite sum of $\frac{a}{1-r} =\frac{-1}{1+1} = -\frac{1}{2}$, though geometric series are usually only considered valid for $|r| < 1$, which is not the case here.
Additionally, note that naive manipulation of the series can lead to values of $-1$ or $0$, in what is called the Eilenberg–Mazur swindle:
$$ -1 + 1-1+1-1+1-\,... =(-1+1)+(-1+1)+(-1+1)+\,...=0+0+0+\,...=0$$
$$ -1 + 1-1+1-1+\,... =-1+(1-1)+(1-1)+\,...=-1+0+0+0+\,...=-1 $$
Thus, the series can have no sum, or sums of $-\frac{1}{2}$, $0$ or $-1$ depending on what axioms you are using and what purpose you are finding an infinite sum for.
