Mathematical induction inequality: $\sum_{r=1}^n\frac{1}{\sqrt r}>\sqrt n,~n\geq 2$ [closed]

Question

I am trying to solve the following problem with mathematical induction but I can't quite seem to work it out.

Prove by mathematical induction that $$\sum_{r=1}^{n}{r^{-\frac12}}>n^{\frac 12}, \forall n\in \Bbb Z, n\ge 2$$

Answer 1

The important observation is that $$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\le \frac1{\sqrt {n+1}} $$

Answer 2

Multiply both sides by $\sqrt{n}$. Then you get $$ \sum_{r=1}^n \sqrt{\frac{n}{r}} \quad \mathrm{vs}\quad n $$ Now observe that in the left side each number is $\ge1$, and there are exactly $n$ terms. Since $n\geq2$ there is a number greater than 1 on the left side. So the sum is greater than $n$.

Answer 3

Base Case: $r=2$

Show that $1^{-\frac12}+2^{-\frac12}>2^\frac12$, that is $1+\frac{1}{\sqrt{2}}>\sqrt{2}$

Then the inductive step: you should assume that for $n=k$ $$\sum_{r=1}^{k}{r^{-\frac 12}}>k^\frac12$$ and show that $k+1$ follows from that, i.e use it to infer: $$\sum_{r=1}^{k+1}{r^{-\frac 12}}>(k+1)^\frac12$$

As a hint: use that: $$\sum_{r=1}^{k+1}{r^{-\frac 12}}=\sum_{r=1}^{k}{r^{-\frac 12}}+(k+1)^{-\frac 12}$$