# Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$ [duplicate]

How do I prove the following? $$\sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}},$$ for all $n \in\mathbb{Z}$, $n\ge 2$.

• Multiply both sides by $\sqrt{n}$? Dec 6, 2011 at 20:53

For any $r \in [1, n]$, we have $$\frac{1}{\sqrt{r}} \geqslant \frac{1}{\sqrt{n}}.$$ with strict inequality for $1 \leqslant r \lt n$. Adding all these $n$ inequalities, $$\sum_{r=1}^{n} \frac{1}{\sqrt{r}} \gt n \cdot \frac{1}{\sqrt{n}} = \sqrt{n}.$$

Proof using induction. The OP requested a proof using induction. I am assuming you can handle the base case $n=2$.

Now, assume that the inequality holds for some $n \geqslant 2$; we will verify the inequality for $n+1$: \begin{align*} \sum_{r=1}^{n+1} \frac{1}{\sqrt{r}} &= \sum_{r=1}^{n} \frac{1}{\sqrt{r}} + \frac{1}{\sqrt{n+1}} \\ &\gt \sqrt{n} + \frac{1}{\sqrt{n+1}} \\ &= \sqrt{n+1} + \frac{1}{\sqrt{n+1}} - (\sqrt{n+1} - \sqrt{n}) \\ &= \sqrt{n+1} + \frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &\gt \sqrt{n+1} , \end{align*} which is what we want to show.

Notice that out second inequality is a bit too crude. robjohn's answer shows how to get a better bound by being more careful.

• Is there any way to prove it using mathematical induction? Dec 5, 2011 at 23:37
• I am not sure why you would want an induction proof. Even if it were possible at all, it will definitely be only more difficult and less enlightening. Dec 5, 2011 at 23:39
• Yes, but if one does a routine induction the proof will look harder than the nice simple argument given by Srivatsan. I can do one if you will promise not to accept it. Dec 5, 2011 at 23:44
• I have added an induction argument, @John. Dec 5, 2011 at 23:45
• @Srivatsan: with induction, you can get a better bound. However, for the bound requested, there is no reason to use induction.
– robjohn
Dec 5, 2011 at 23:50

We can even do better: $$\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}<\frac{1}{2\sqrt{n-1}}\tag{1}$$ Summing, we get $$\sqrt{n}-1<\frac{1}{2}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n-1}}\right)\tag{2}$$ So, for $n\ge2$, $$\sqrt{n}<1+\frac{1}{2}\left(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n-1}}\right)\tag{3}$$ which is a better bound for every $n$.