I still do not have the answers I want for this thread I made mathematical induction methods, or rather it made me more confused in what I potentially am doing wrong. So I will try to ask again and see what is wrong with the method (I do know how to do it the standard way, just wondering if this method also works):
Question: Prove by mathematical induction that for $n\ge2, n\in \mathbb{Z^+}$: $$\sum_{r=1}^{n}\frac{1}{\sqrt{r}}>\sqrt{n}$$
Method: If I go straight to the induction step to proving $P(k+1)$, then we have: $$\sum_{r=1}^{k+1}\frac{1}{\sqrt{r}}>\sqrt{k+1}$$ $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}$$ $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}>\sqrt{k}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}=\frac{1}{\sqrt{k+1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}}$$ However $k\ge 2$, and therefore the expression: $$\frac{1}{\sqrt{k+1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}}>0$$ Hence: $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}>\frac{1}{\sqrt{k+1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}}>0$$ $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}-\sqrt{k+1}>0$$ $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}>\sqrt{k+1}$$
Hence by mathematical induction, we proved the result.
Is this a valid method?