Proving $\forall n \in \mathbb{N}^*, 2\sqrt{n+1}-2 \le S_n \le \sqrt{n}+\sqrt{n-1}$

Question

Suppose the sequence $S_n=\displaystyle \Sigma_{k=1}^n\frac{1}{\sqrt{k}}$.

How can I prove that:

$$\forall n \in \mathbb{N}^*, 2\sqrt{n+1}-2 \le S_n \le \sqrt{n}+\sqrt{n-1}$$

Answer 1

How? By induction over $n\geqslant1$.

Answer 2

To elaborate on Did's answer, which you think is a comment, show that

$$ 2 \sqrt{n+2} - 2 \sqrt{n+1} \leq \frac{1}{\sqrt{n+1} } \leq \sqrt{n+1} - \sqrt{n-1}$$

I'd show you one side, say we do the RHS. Multiplying by $\sqrt{n+1}$, it suffices to show that

$$ 1 \leq n+1 - \sqrt{n^2-1},$$

or that

$$\sqrt{n^2-1} \leq n,$$

which is obviously true.

Now you do the LHS.