Show that $1/\sqrt{1} + 1/\sqrt{2} + … + 1/\sqrt{n} \leq 2\sqrt{n}-1$

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  • How to prove the inequality $2\sqrt{n + 1} − 2 \le 1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n} \le 2\sqrt n − 1$?

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Solutions Collecting From Web of "Show that $1/\sqrt{1} + 1/\sqrt{2} + … + 1/\sqrt{n} \leq 2\sqrt{n}-1$"

hint: $\dfrac{1}{\sqrt{k}}<2\sqrt{k}-2\sqrt{k-1}$. You let $k$ runs from $1$ to $n$, and add the inequalities up.

If the proposition holds true for $n=m$

$$\sum_{r=1}^m\dfrac1{\sqrt r}\le2\sqrt m-1$$

$$\implies\sum_{r=1}^{m+1}\dfrac1{\sqrt r}=\dfrac1{\sqrt{m+1}}+\sum_{r=1}^m\dfrac1{\sqrt r}\le\dfrac1{\sqrt{m+1}}+2\sqrt m-1$$

It if sufficient to show $$\dfrac1{\sqrt{m+1}}+2\sqrt m-1\le2\sqrt{m+1}-1$$

$$\iff\dfrac1{\sqrt{m+1}}\le2[\sqrt{m+1}-\sqrt m]=\dfrac2{\sqrt{m+1}+\sqrt m}$$

$$\iff \sqrt{m+1}\ge\sqrt m$$ which is true

Without induction you could use this inequality

$$\frac1{\sqrt k}\le2(\sqrt{k}-\sqrt{k-1})$$
and then you telescope. Notice also that the same inequality is useful for your attempt.