Thoughts about the convergence (conditional or absolute) and divergence of series:
I'm trying to prove that these series converge conditionally, converge absolutely, or diverge.
$\sum^∞_\mathrm{n=0}(−1)^n\frac{1}{n!}$
My initial thought was that the sequence must converge to $0$. The sign of each term in the sequence of partial sums will depend entirely on $(-1)^n$ as $n \geq 0$.
We know $0!$ is $1$, hence the sequence of partial sums has $1$ as its first term. Each subsequent term will be smaller than the last, because $\frac{1}{n!} > \frac{1}{(n+1)!}$. I see the overall sequence converging as two subsequences approaching $0$ from above and below [$n=1$ gives $(-1)*1$, $n=2$ gives $(-1)^2*\frac{1}{2*1} = \frac{1}{2}$].
Say you had another series. This time:
$\sum^∞_\mathrm{n=0}(−1)^n\frac{1}{(2n-1)}$
By the logic for the first series, this series would also converge. Momentarily ignoring the sign given by $(-1)^n$, each term $\frac{1}{(2n-1)}$ would be greater than the subsequent term $\frac{1}{(2(n+1) - 1)}$. Once you apply the $(-1)^n$, the sequence would also approach 0 from above and below.
I don't know how to wrap these thoughts up into a more formal proof. Am I looking at this the right way?