What is the limit of $\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}$?

There is a telescopic sum in disguise. Since: $$\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}=\frac{1}{2}\left(\sqrt{2k+1}-\sqrt{2k-1}\right)\tag{1}$$ by summing $(1)$ for $k$ that goes from $1$ to $n$ we have: $$ \sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}} = \frac{1}{2}\left(\sqrt{2n+1}-\sqrt{1}\right)\tag{2}$$ hence by multiplying both sides by $\frac{1}{\sqrt{n}}$ and taking the limit as $n\to +\infty$ we clearly have: $$ \lim_{n\to +\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}=\color{red}{\frac{1}{\sqrt{2}}}.\tag{3}$$

Hint 1:
$\frac{1}{\sqrt1+\sqrt3}=\frac{\sqrt3-\sqrt1}{2}$

$\frac{1}{\sqrt3+\sqrt5}=\frac{\sqrt5-\sqrt3}{2}$

and so on, so you are left with $\frac{\sqrt{2n+1}-1}{2}$ inside the brackets,
hope you can do it from here now.

Hint 2: (see only if you need it)

After multiplying by $\frac{1}{\sqrt{n}}$ it becomes $\frac{\sqrt{2+\frac{1}{n}}-\frac{1}{\sqrt{n}}}{2}. $ Now what is $\frac 1n$ and $\frac{1}{\sqrt{n}}$ if $n \to{+ \infty}$?

Other hint.

For $n \in \mathbb N$ and $x \in [n,n+1]$ you have $$\frac{1}{2\sqrt{2x+1}} \le \frac{1}{\sqrt{2n-1}+\sqrt{2n+1}} \le \frac{1}{2\sqrt{2x-3}}$$ which enables to compare the sum with integrals.