How prove this inequality? show that
$$\dfrac{\sqrt{2}}{2}<f(n)=\dfrac{\sqrt{2n+1}-1}{1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{n}}}<\dfrac{\sqrt{3}}{2}$$
I know this
$$\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}<\dfrac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$$
By the way I can use Stolz lemma find the
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{\dfrac{1}{\sqrt{n}}}=\dfrac{\sqrt{2}}{2}$$
but this can't prove this inequality,Thank you

The estimate $1/\sqrt{n} < 2(\sqrt{n}-\sqrt{n-1})$ yields
$$\sum_{k=1}^n \frac{1}{\sqrt{k}} < 1 + \sum_{k=2}^n 2(\sqrt{k}-\sqrt{k-1}) = 2\sqrt{n} - 1,$$
and so
$$\frac{\sqrt{2n+1}-1}{\sum_{k=1}^n \frac{1}{\sqrt{k}}} > \frac{\sqrt{2n+1}-1}{2\sqrt{n}-1} > \frac{\sqrt{2}}{2}.$$
For an upper bound, we could look at the analogous $1/\sqrt{n} > 2(\sqrt{n+1}-\sqrt{n})$, which yields
$$\sum_{k=1}^n \frac{1}{\sqrt{k}} > 2(\sqrt{n+1}-1).$$
That would leave
$$\frac{\sqrt{2n+1}-1}{\sqrt{n+1}-1} < \sqrt{3}$$
to be shown, or
$$\sqrt{3}-1 < \sqrt{3(n+1)}-\sqrt{2n+1} = \frac{n+2}{\sqrt{3(n+1)}+\sqrt{2n+1}}.$$
A trivial estimate shows the right hand side is $> \frac{\sqrt{n+2}}{\sqrt{3}+\sqrt{2}}$, which yields the inequality for $n \geqslant 7$.
The cases $1 \leqslant n \leqslant 6$ can be verified by hand.
