One of the possible answers.
$$ \frac{1}{2} < \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} < 1 \;\;\;(1)$$
Since $\displaystyle H_m = \sum_{n=1}^{m}\frac{1}{n}$ is increasing, if it is bounded, it converges, otherwise diverges to infinity.
Assume it converges, then:
$$\sum_{n=1}^{\infty}\frac{1}{n} = 2\frac{1}{2}+2\frac{1}{4}+2\frac{1}{6}+...+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = \sum_{n=1}^{\infty}\frac{1}{n} + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$$
which is giving
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=0$$
and that is not the case.
For $\displaystyle (1)$, the left side is obvious, the right side
$$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = \sum_{n=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n} = \sum_{n=1}^{\infty}\frac{1}{4n^2-2n} $$ $$< \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}<\frac{1}{2} + \frac{1}{2}\sum_{n=2}^{\infty} \left (\frac{1}{n-1}-\frac{1}{n} \right ) $$
$$= \frac{1}{2} + \frac{1}{2} + \frac{1}{2}\sum_{n=2}^{\infty} \left (-\frac{1}{n}+\frac{1}{n} \right ) = 1$$
by canceling.