<s>[INST] 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.
 [/INST] 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.
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