<s>[INST] Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}<2\sqrt{n}+\dfrac{1}{\sqrt{n+1}}=$
$2\sqrt{n}+\dfrac{2}{2\sqrt{n+1}}<2\sqrt{n}+\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\sqrt{n}+2(\sqrt{n+1}-\sqrt{n})=2\sqrt{n+1}$.
However, despite hours of trying, I cannot figure out how to show that $2(\sqrt{n+1}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}$.
 [/INST] Note: I can see that there are already fine answers using induction!
An ad in favor of graph paper. The outcome of the standard sum/integral comparison is
$$  2 \sqrt {n+1} - 2 \;  < \; 1  + \frac{1}{\sqrt 2 }  + \frac{1}{\sqrt 3 } + \cdots + \frac{1}{\sqrt n } \; < \; 2 \sqrt n - 1 $$

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