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?
 A: 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.
A: 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:
$$ \lim_{n\to +\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}=\color{red}{\frac{1}{\sqrt{2}}}.\tag{3}$$
A: Notice, we have $$\lim_{n\to \infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 1+\sqrt 3}+\frac{1}{\sqrt 3+\sqrt 5}+\cdots +\frac{1}{\sqrt {2n-1}+\sqrt {2n+1}}\right)$$
$$=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt {2r-1}+\sqrt {2r+1}}\right)$$
$$=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt n}\left(\frac{\sqrt {2r+1}-\sqrt {2r-1}}{2r+1-(2r-1)}\right)$$
$$=\frac{1}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt {2r+1}-\sqrt {2r-1}}{\sqrt n}$$
$$=\frac{1}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt{2r}\left(1+\frac{1}{2r}\right)^{1/2}-\sqrt{2r}\left(1-\frac{1}{2r}\right)^{1/2}}{\sqrt n}$$
Using binomial expansion of $\left(1-\frac{1}{2r}\right)^{1/2}$ & neglecting higher power terms as $\left(\frac{1}{2r}\right)^2$, $\left(\frac{1}{2r}\right)^3,\ldots $
$$=\frac{\sqrt 2}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt r}{\sqrt n} \left[\left(1+\frac{1}{2}\frac{1}{2r}\right)-\left(1-\frac{1}{2}\frac{1}{2r}\right)\right]$$
$$=\frac{1}{\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt r}{\sqrt n} \left[\frac{1}{2r}\right]$$
$$=\frac{1}{2\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt {nr}}$$
$$=\frac{1}{2\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{\sqrt {\frac{r}{n}}}$$
$$=\frac{1}{2\sqrt 2}\int_{0}^{1}\frac{dx}{\sqrt {x}}$$
$$=\frac{1}{2\sqrt 2}[2\sqrt x]_{0}^{1}$$
$$=\frac{1}{2\sqrt 2}[2-0]$$
$$=\frac{1}{\sqrt 2}$$
A: 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 $\frac{1}{\sqrt{2}}$.
A: 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 $n \to{+ \infty}$?

