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

$$\lim_{n\to \infty} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)=?$$

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?

Answer

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:
$$\lim_{n\to +\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}=\color{red}{\frac{1}{\sqrt{2}}}.\tag{3}$$