Find, analytically, the value of the following limit. How would one prove that$$\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k-1}}$$ converges (rather slowly) to $\frac {1}{\sqrt{2}}$, which appears obvious from numerical computation.
 A: $$\frac{1}{\sqrt{2k}+\sqrt{2k-1}}=\frac{1}{\sqrt2\left(\sqrt{k}+\sqrt{k-\frac{1}{2}}\right)}<\frac{1}{\sqrt2\left(\sqrt{k}+\sqrt{k-1}\right)}=\frac{1}{\sqrt2}\left(\sqrt{k}-\sqrt{k-1}\right),$$
$$\frac{1}{\sqrt{2k}+\sqrt{2k-1}}=\frac{1}{\sqrt2\left(\sqrt{k}+\sqrt{k-\frac{1}{2}}\right)}>\frac{1}{\sqrt2\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{1}{\sqrt2}\left(\sqrt{k+1}-\sqrt{k}\right)$$ and use the telescoping summation.
A: Well, we can write the sum under limit as $$\frac{1}{n}\sum_{k=1}^{n}\dfrac{1}{\sqrt{2}\left(\sqrt{\dfrac{k}{n}}+\sqrt{\dfrac{k}{n}-\dfrac{1}{2n}}\right)}$$ which is a Riemann sum for $f(x) = 1/(2\sqrt{2x})$ over $[0,1]$ with partition points $x_k=k/n$ and suitable tag points $t_k$. The desired limit is $2^{-1/2}\int_{0}^{1}\frac{dx}{2\sqrt{x}}=1/\sqrt{2}$.
A: Another interesting way to prove the lower bound.
Consider:
$$A=\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k-1}}$$
$$I_k=\frac{1}{\sqrt{2k}+\sqrt{2k-1}}=\frac{2}{\sqrt{\pi}}\int_0^\infty e^{-(4k-1)x^2} e^{-4k x^2 \sqrt{1-1/(2k)}} dx$$
Since $(\sqrt{2k}+\sqrt{2k-1})^2=4 k-1+4 \sqrt{k(k-1/2)}$.
$$\sqrt{1-\frac{1}{2k}}= 1-\frac{1}{4k}-\frac{1}{32k^2}-O \left( \frac{1}{8k^3} \right)<1-\frac{1}{4k}$$
We have:
$$I_k > \frac{2}{\sqrt{\pi}} \int_0^\infty e^{-8kx^2} dx$$
$$A \geq \frac{2}{\sqrt{\pi}} \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n  \int_0^\infty e^{-8kx^2} dx= \\ =\frac{2}{\sqrt{\pi}} \lim_{n\to\infty}\frac{1}{\sqrt{n}} \int_0^\infty \frac{e^{-8 x^2}-e^{-8(n+1) x^2}}{1-e^{-8 x^2}}  dx$$
$$A \geq \frac{2}{\sqrt{\pi}} \lim_{n\to\infty}\frac{1}{n} \int_0^\infty \frac{e^{-8 y^2/n}-e^{-8(1+1/n) y^2}}{1-e^{-8 y^2/n}}  dy$$
$$A \geq \frac{2}{\sqrt{\pi}} \lim_{n\to\infty}\frac{1}{n} \int_0^\infty \frac{1-e^{-8 y^2}}{e^{8 y^2/n}-1}  dy=\frac{1}{4 \sqrt{\pi}}\int_0^\infty \frac{1-e^{-8 y^2}}{y^2}  dy$$

$$A \geq \frac{1}{\sqrt{2 \pi}}\int_0^\infty \frac{1-e^{-z^2}}{z^2}  dz= \frac{1}{\sqrt{2}}$$

The last integral is easy to prove using integration by parts.
