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

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

• You should show what you have tried. Where specifically are you stuck? Have you done the base case at least?
– Dave
Commented Jun 4, 2018 at 18:11
• I am fine with the basis step, it is the inductive step that is not making sense; where I am required to show that the statement is true for the parameter value n = k + 1. Commented Jun 4, 2018 at 18:22
• For other ways than induction, the reference post is math.stackexchange.com/questions/2149448/…
– zwim
Commented Jun 4, 2018 at 18:52

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

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 $n$.

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: $$\sum_{r=1}^{k+1}{r^{-\frac 12}}=\sum_{r=1}^{k}{r^{-\frac 12}}+(k+1)^{-\frac 12}$$