<s>[INST] Help verifying proof to proving that for all natural numbers $\sqrt{1} \leq$ than the sum Prove that for all natural numbers $n$,
$$\sqrt{n} \le 1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}.$$
Solution: We must prove that $1 + \frac 1 {\sqrt{2}}, + \frac 1 {\sqrt{3}} +\,\cdots\, + \frac 1 {\sqrt{k+1}} \geq {\sqrt{k+1}}$
Add $\sqrt{k+1}-\sqrt{k}$ to both sides of first inequality, we get:
$1+\frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \sqrt{k+1}-\sqrt{k} \geq \sqrt{k+1}$
However,
$$\sqrt{k+1}-\sqrt{k}=\frac{(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k})}{(\sqrt{k+1}+\sqrt{k})}=\frac{k+1-k}{(\sqrt{k+1}+\sqrt{k})}=\frac{1}{(\sqrt{k+1}+\sqrt{k})} \leq \frac{1}{\sqrt{k+1}}$$
So:
$\sqrt{k+1} \leq 1 + \frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}} + \cdots + \frac{1}{\sqrt{k}} + \sqrt{k+1}-\sqrt{k} \leq 1+\frac{1}{\sqrt{2}} + \frac 1 {\sqrt{3}}+\cdots + \frac{1}{\sqrt{k+1}}$
 [/INST] You are right. The inductive proof needs to establish the link between
$$\sqrt n\le1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}$$
 and
$$\sqrt{n+1}-\frac1{\sqrt{n+1}}\le1\,+\,\frac{1}{\sqrt{2}}\,+\,\frac{1}{\sqrt{3}}\,+\,\cdots\,+\,\frac{1}{\sqrt{n}}.$$
Obviously,
$$\sqrt{n+1}-\frac1{\sqrt{n+1}}=\frac n{\sqrt{n+1}}<\sqrt n$$ does the trick.

For a better estimate, use
$$\frac1{\sqrt{x+1}}<\frac1{\sqrt{\lfloor x\rfloor}}\le\frac1{\sqrt x},$$ then by integration
$$\int_1^{n+1}\frac{dx}{\sqrt{x+1}}<\sum_{k=1}^n\frac1{\sqrt k}\le\int_1^{n+1}\frac{dx}{\sqrt x},$$ 
$$2(\sqrt{n+2}-\sqrt2)<\sum_{k=1}^n\frac1{\sqrt k}\le2(\sqrt{n+1}-1).$$ 
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