Comparing asymptotics to alternate series test I am a bit confused when applying asymptotics:
Let's consider $\sum\limits_{k=1}^\infty(-1)^k\frac{1}{k}$. We know that it is convergent (alternate series test). However, if I apply asymptotics I get the following:
$$
(-1)^k\frac{1}{k}= O\left(\frac{1}{k}\right)\implies \sum\limits_{k=1}^\infty(-1)^k\frac{1}{k}=\sum\limits_{k=1}^\infty O\left(\frac{1}{k}\right).
$$
The resulting series is no longer convergent although it has the same asymptotic behavior?! How does this fit together?
 A: Let's say we have two series $$\sum_{k=1}^{\infty}a_{k}\quad \text{and}\quad \sum_{k=1}^{\infty}b_{k}$$
and we know that $$a_{k} = \mathcal{O}(b_{k})\quad \text{as} \quad k\to\infty.$$
Remember that this means that there is some constant $M>0$ and some index $n$ such that $$|a_{k}|\leq M|b_{k}| \quad\text{for all } k\geq n.$$
Now, this is useful for determining convergence of $\sum_{k=1}^{\infty}a_{k}$, but only in the case that we already know that $\sum_{k=1}^{\infty}b_{k}$ converges absolutely.  Indeed, if $\sum_{k=1}^{\infty}b_{k}$ converges then we can say that
$$\sum_{k=1}^{\infty}|a_{k}| = \sum_{k=1}^{n-1}|a_{k}| + \sum_{k=n}^{\infty}|a_{k}| \leq \sum_{k=1}^{n-1}|a_{k}| + M\sum_{k=n}^{\infty}|b_{k}| $$ which converges because $\sum_{k=n}^{\infty}|b_{k}|$ is the tail of a convergent series.  However, if $\sum_{k=1}^{\infty}|b_{k}|$ diverges, then all we've done is bound $\sum_{k=1}^{\infty}a_{k}$ above by a divergent series, which we know from the comparison test doesn't tell us anything about the convergence of $\sum_{k=1}^{\infty}a_{k}$.
