What is the limit of $\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}$? $$\lim_{n\to \infty} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)=? $$
I tried with the squeeze theorem, though I got the upper bound, but I couldn't find the lower bound. I also tried to solve it with the order limit theorem, but without any success. I guessed the result should be $\frac{1}{\sqrt{2}}$. How can I do this?

If you don't spot that the series telescopes (and this is always worth a check - usually careful inspection of the first few terms will suffice) then here is a solution using the Stolz–Cesàro theorem.
Write the desired series as a fraction:
$$S_n = \frac{\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}}{\sqrt{n}} = \frac{a_n}{b_n} $$
Now $b_n = \sqrt{n}$ is strictly monotone and divergent (since it is strictly increasing and $b_n \to +\infty$). We need this condition for Stolz–Cesàro to apply.
Now consider the "fraction of differences" (does it have a more technical name?):
$$\frac{a_n - a_{n-1}}{b_n - b_{n-1}} = \frac{\left(\sqrt{2n+1}+\sqrt{2n-1}\right)^{-1}}{\sqrt{n}-\sqrt{n-1}} = \frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac{1+\sqrt{1-\frac{1}{n}}}{\sqrt{2+\frac{1}{n}}+\sqrt{2-\frac{1}{n}}}$$
The numerator $1+\sqrt{1-\frac{1}{n}} \to 2$ and the denominator $\sqrt{2+\frac{1}{n}}+\sqrt{2-\frac{1}{n}} \to 2\sqrt{2}$ so the fraction tends to $\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$.
Then since $b_n$ was strictly monotone and divergent, and the limit of $\frac{a_n - a_{n-1}}{b_n - b_{n-1}}$ exists and equals $\frac{1}{\sqrt{2}}$, we can conclude from the Stolz–Cesàro theorem that the limit of $\frac{a_n}{b_n}$ also exists and also equals $\frac{1}{\sqrt{2}}$.
