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

Solution 1:

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.

Solution 2:

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 $n$.