\begin{aligned} \mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1=\frac{\left(-1\right)^{n}}{n}\int_{0}^{1}{\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}x}\,\mathrm{d}x}&=\frac{\left(-1\right)^{n}}{n}+\frac{\left(-1\right)^{n}}{n}\int_{0}^{1}{\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}x}-1\right)\mathrm{d}x}\\ &=\frac{\left(-1\right)^{n}}{n}+\frac{1}{n^{2}}\int_{0}^{1}{\int_{0}^{1}{x\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}xy}\,\mathrm{d}y}\,\mathrm{d}x}\\ \mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1 &=\frac{\left(-1\right)^{n}}{n}+v_{n} \end{aligned}
With $ v_{n}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n^{2}}\right) \cdot $
From that remains the following : \begin{aligned} \left|\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1\right|&\geq\left|\left|\frac{\left(-1\right)^{n}}{n}\right|-\left|v_{n}\right|\right|=\frac{1}{n}-\frac{1}{n^{2}}\int_{0}^{1}{\int_{0}^{1}{x\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}xy}\,\mathrm{d}y}\,\mathrm{d}x}\end{aligned}
Since $ \sum\limits_{n\geq 1}{\frac{1}{n}} $ diverges, $ \sum\limits_{n\geq 1}{\left(\frac{1}{n}-\frac{1}{n^{2}}\int_{0}^{1}{\int_{0}^{1}{x\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}xy}\,\mathrm{d}y}\,\mathrm{d}x}\right)} $ will diverge, and thus, by comparaison, $ \sum\limits_{n\geq 1}{\left|\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1\right|} $ diverges.
Now since $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n}}{n}} $ converges by the AST,
and $ \sum\limits_{n\geq 1}{v_{n}} $ converges by comparaison test, we have that $ \sum\limits_{n\geq 1}{\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1\right)} $ converges.
Now, we can do pretty much the same thing for the second one :
\begin{aligned} \mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}}-1=\frac{\left(-1\right)^{n}}{\sqrt{n}}\int_{0}^{1}{\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}x}\,\mathrm{d}x}&=\frac{\left(-1\right)^{\sqrt{n}}}{n}+\frac{\left(-1\right)^{n}}{\sqrt{n}}\int_{0}^{1}{\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}x}-1\right)\mathrm{d}x}\\ &=\frac{\left(-1\right)^{n}}{\sqrt{n}}+\frac{1}{n}\int_{0}^{1}{\int_{0}^{1}{x\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}xy}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\frac{\left(-1\right)^{n}}{\sqrt{n}}+\frac{1}{2n}+\frac{1}{n}\int_{0}^{1}{\int_{0}^{1}{x\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}xy}-1\right)\mathrm{d}y}\,\mathrm{d}x}\\ &=\frac{\left(-1\right)^{n}}{\sqrt{n}}+\frac{1}{2n}+\frac{\left(-1\right)^{n}}{n\sqrt{n}}\int_{0}^{1}{\int_{0}^{1}{\int_{0}^{1}{x^{2}y\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}xyt}\mathrm{d}t}\,\mathrm{d}y}\,\mathrm{d}x}\\ \mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1 &=\frac{\left(-1\right)^{n}}{\sqrt{n}}+\frac{1}{2n}+w_{n} \end{aligned}
With $ w_{n}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n\sqrt{n}}\right) \cdot $
We have that $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n}}{\sqrt{n}}} $ converges by the AST,
and $ \sum\limits_{n\geq 1}{w_{n}} $ converges by comparaison test, but with $ \sum\limits_{n\geq 1}{\frac{1}{n}} $ being divergent. We have that $ \sum\limits_{n\geq 1}{\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}}-1\right)} $ diverges.
And thus, $ \sum\limits_{n\geq 1}{\left|\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}}-1\right|} $ also diverges.