<s>[INST] Prove that the series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}$ is convergent by Cauchy Criterion. Prove that the series 
$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-\cdots$ is convergent by Cauchy Criterion
Let $\epsilon >0$. We will find $N \in \mathbb{N}$ such that when $n>m \geq N$ then $|s_n-s_m|< \epsilon$. Where $|s_n-s_m|=|a_{m+1}-a_{m+2}+a_{m+3}- \cdots \pm a_n|$. Note that $(\frac{1}{\sqrt{m+1}} - \frac{1}{\sqrt{m+2}}+\frac{1}{\sqrt{m+3}}-\frac{1}{\sqrt{m+4}} \cdots + \frac{1}{{\sqrt{n}}}or +(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}) >0$ 
Also note we have that $\frac{1}{\sqrt{m+1}} -(\frac{1}{\sqrt{m+2}}-\frac{1}{\sqrt{m+3}})-(\frac{1}{\sqrt{m+4}}-\frac{1}{\sqrt{m+5}})\cdots -\frac{1}{\sqrt{n}}or-(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}})$ 
So $0< \frac{1}{\sqrt{m+1}} -\frac{1}{\sqrt{m+2}}+ \cdots \frac{1}{\sqrt{n}}<\frac{1}{\sqrt{m+1}}<\frac{1}{\sqrt{m}}$ Then as soon as $\frac{1}{\sqrt{m}}< \epsilon$ we will have that $\frac{1}{\sqrt{m+1}} -\frac{1}{\sqrt{m+2}}+ \cdots \frac{1}{\sqrt{n}}< \epsilon$. Thus choose $N \in \mathbb{N}$, where $N> \frac{1}{\epsilon^2}$. Therefore if $n>m\geq N$ then we know that $|a_{m+1}-a_{m+2}+a_{m+3}- \cdots \pm a_n|< \frac{1}{\sqrt{m}}\leq \frac{1}{\sqrt{N}} < \epsilon$.                 
 [/INST] Note that the partial sums alternately overshoot and undershoot the limit.  
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