Prove by induction question

Question

Please help me to prove this:$$\sum_{r=1}^n \frac{1}{\sqrt r} > \sqrt n$$ for $$n \ge 2, n \in \Bbb Z$$ I finished the first step with this: $$\frac {2+\sqrt 2}{2}> \sqrt 2 \rightarrow True$$ But I am not sure whether this is correct. Then I don't know if it is fine to leave the 2nd step like this: $$\frac {1}{\sqrt 1} + \frac {1}{\sqrt 2} + \cdots + \frac {1}{\sqrt k} > \sqrt k \rightarrow True$$ And finally in the 3rd step I am stuck with this and I don't know how to move on: $$\sum_{r=1}^{k+1} \frac{1}{\sqrt r} = \sum_{r=1}^{k} \frac{1}{\sqrt r} + \frac {1}{\sqrt {k+1}}$$ Thanks for your help in advance.

Answer 1

You had the general idea, but it seems that you're not quite clear on how to do a proof by induction.

You were right that you first start with $n=2$. $$\sum_{r=1}^2 \frac{1}{\sqrt{r}} = 1 + \frac{1}{\sqrt{2}}$$ $$\sqrt{2} >1 \implies \sqrt{2}+1 > 2 \implies 1 + \frac{1}{\sqrt{2}} > \sqrt{2}$$ Thus we've shown that it holds for $n=2$. Here's the main induction step. Suppose that $$\sum_{r=1}^k \frac{1}{\sqrt{r}} > \sqrt{k}$$ Then $$\sum_{r=1}^{k+1} \frac{1}{\sqrt{r}} = \sum_{r=1}^{k} \frac{1}{\sqrt{r}} +\frac{1}{\sqrt{k+1}}$$ By our assumption $$\sum_{r=1}^{k} \frac{1}{\sqrt{r}} +\frac{1}{\sqrt{k+1}} > \sqrt{k} + \frac{1}{\sqrt{k+1}}$$ Now we'll show that the above is bigger than $\sqrt{k+1}$. It's clear that $$\sqrt{k}\sqrt{k+1} > k$$ Thus $$\begin{align} \sqrt{k}\sqrt{k+1}+1 &> k+1\\ \sqrt{k}+\frac{1}{\sqrt{k+1}} &> \sqrt{k+1}\\ \end{align}$$

Answer 2

You need to show that

$$\sum_{r=1}^n \frac{1}{\sqrt r} > \sqrt n\implies \sum_{r=1}^n\frac{1}{\sqrt r} +\frac1{\sqrt{n+1}} > \sqrt{n+1}.$$

We have

$$\sum_{r=1}^n \frac{1}{\sqrt r} > \sqrt n\implies \sum_{r=1}^n\frac{1}{\sqrt r} +\frac1{\sqrt{n+1}} >\sqrt n+\frac1{\sqrt{n+1}},$$ and

$$\sqrt n+\frac1{\sqrt{n+1}}>\sqrt{n+1}$$ because

$$\sqrt n\sqrt{n+1}+1>n+1.$$

This proves the claim.