I want some help, please, to solve this problem.
Let $$\forall n\in\mathbb{N}^*,\ S_n=\sum_{k=1}^n\frac{1}{\sqrt{k}}$$
I want to prove that $\forall n\in\mathbb{N}^*,\ S_n\leqslant\sqrt{n-1}+\sqrt{n}$
and that $\forall n\in\mathbb{N}^*,\ S_n\geqslant2\sqrt{n+1}-2$
I used the Mathematical Induction method, so for $n=1$ we have : $S_1=1$ and $\sqrt{1-1}+\sqrt{1}=1$
Then the statement in this case is satisfied. Next we assume another real non-null number $k/\ S_k\leqslant\sqrt{k-1}+\sqrt{k}$ , and we prove that $S_{k+1}\leqslant\sqrt{k}+\sqrt{k+1}$ , but I'm stuck.
So please help me if you can. Thanks!
We have
$$S_{k+1}=S_k+\frac1{\sqrt{k+1}}\stackrel{?}\le \sqrt{k}+\sqrt{k+1}$$
which is true indeed
$$S_k+\frac1{\sqrt{k+1}}\le \sqrt{k-1}+\sqrt{k}+\frac1{\sqrt{k+1}}\le \sqrt{k}+\sqrt{k+1}$$
since
$$\sqrt{k-1}+\sqrt{k}+\frac1{\sqrt{k+1}}\le \sqrt{k}+\sqrt{k+1}\iff \sqrt{k-1}+\frac1{\sqrt{k+1}}\le \sqrt{k+1}$$
$$\sqrt{k^2-1}+1\le k+1 \iff \sqrt{k^2-1}\le k \iff k^2-1 \le k^2 \iff -1\le 0$$
Edit
We also need to prove that
$$S_{k+1}=S_k+\frac1{\sqrt{k+1}}\stackrel{?}\ge 2\sqrt{k+2}-2$$
which is true indeed
$$S_k+\frac1{\sqrt{k+1}}\ge 2\sqrt{k+1}-2+\frac1{\sqrt{k+1}}\ge 2\sqrt{k+2}-2$$
since
$$2\sqrt{k+1}-2+\frac1{\sqrt{k+1}}\ge 2\sqrt{k+2}-2\iff 2\sqrt{k+1}+\frac1{\sqrt{k+1}}\ge 2\sqrt{k+2}$$
$$2(k+1)+1\ge 2\sqrt{(k+2)(k+1)} \iff 2k+3 \ge 2\sqrt{(k+2)(k+1)}$$
$$4k^2+12k+9 \ge 4k^2+12k+8 \iff 1\ge 0$$