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Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction

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

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    Multiply both sides by $\sqrt{n}$?2011-12-06

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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.

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    @Srivatsan: with induction, you can get a better bound. However, for the bound requested, there is no reason to use induction.2011-12-05
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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 $n$.