Proove by the Complete Induction for every $n\in \mathbb{N}, n \geq 1$ $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} > \sqrt{n}$$.
I know only the basis - $\frac{1}{\sqrt{1}} \geq \sqrt{1}$.
But, I dont know how to do the change from the regular induction to the induction of Set theroy(Complete Induction).
I don't know about the induction of set theory but the inductive step for ordinary induction goes like this:
Assume
\begin{equation} \frac{1}{\sqrt{1}}+ \frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}>\sqrt{n} \end{equation}
\begin{equation} \sqrt{n}=\frac{n}{\sqrt{n}}>\frac{n}{\sqrt{n+1}} \end{equation} Therefore \begin{equation} \sqrt{n}>\sqrt{n+1}-\frac{1}{\sqrt{n+1}} \end{equation} so \begin{equation} \sqrt{n}+\frac{1}{\sqrt{n+1}}>\sqrt{n+1} \end{equation}
Thus \begin{equation} \frac{1}{\sqrt{1}}+ \frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}>\sqrt{n+1} \end{equation}