Prove that $\sqrt{n} \le \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1$ is true for $n \in \mathbb{N}^{\ge 1}$ I'm trying to solve these induction exercises proposed by the department of mathematics of Oxford University. I don't know how to give a valid proof for the third one which says the following:

Prove that for $n \in \mathbb{N}^{\ge 1}$: $$ \sqrt{n} \le
 \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1 $$

My first approach has been to convert the summation into a function in terms of $n$ so that I could find the maximum and minimum values in the interval. However, there doesn't seem to be an easy way to express the summation as a function. Thanks in advance!

Since
$$
\sqrt{k+1}-\sqrt{k}=\frac1{\sqrt{k+1}+\sqrt{k}}\tag{1}
$$
we have
$$
\frac1{2\sqrt{k+1}}\le\sqrt{k+1}-\sqrt{k}\le\frac1{2\sqrt{k}}\tag{2}
$$
Summing $(2)$ yields
$$
\sum_{k=2}^n\frac1{2\sqrt{k}}\le\sqrt{n}-1\le\sum_{k=1}^{n-1}\frac1{2\sqrt{k}}\tag{3}
$$
which implies
$$
\sqrt{n}-1+\frac1{2\sqrt{n}}\le\sum_{k=1}^n\frac1{2\sqrt{k}}\le\sqrt{n}-\frac12\tag{4}
$$
Therefore,
$$
2\sqrt{n}-2+\frac1{\sqrt{n}}\le\sum_{k=1}^n\frac1{\sqrt{k}}\le2\sqrt{n}-1\tag{5}
$$
Combining $(5)$ and
$$
\sqrt{n}\le\sqrt{n}+\left(n^{1/4}-n^{-1/4}\right)^2=2\sqrt{n}-2+\frac1{\sqrt{n}}\tag{6}
$$
yields
$$
\sqrt{n}\le\sum_{k=1}^n\frac1{\sqrt{k}}\le2\sqrt{n}-1\tag{7}
$$
