<s>[INST] Show that $ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$ Show that $$ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$$
LHS : $ \lim_{n\to\infty}\frac{1}{n}[\frac{n}{\sqrt n}+\frac{n}{\sqrt {n+1}}+\frac{n}{\sqrt {n+2}}.......\frac{n}{\sqrt {2n}}] = \infty$
Let $a_n=\frac{n}{\sqrt{2n}}$ then $a_n = \frac{1}{\sqrt 2}\sqrt{n}$
hence $\lim_{n\to\infty}a_n = \infty$
I continue the problem therefore using Cauchy's first theorem on limits.
But the way I took $a_n$ is it correct?  
Does it not make $a_1 = \frac{1}{\sqrt{2}}$
If my way of taking $a_n$ is wrong, please suggest me the right way. Thank you
 [/INST] You could have quite good approximations using generalized harmonic numbers
$$S_n=\sum_{i=0}^n \frac 1 {\sqrt{n+i}}=H_{2 n}^{\left(\frac{1}{2}\right)}-H_{n-1}^{\left(\frac{1}{2}\right)}$$
Using the asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{p}\right)^{1/2}-\frac{1}{24}
   \left(\frac{1}{p}\right)^{3/2}+O\left(\frac{1}{p^{7/2}}\right)$$ Using it twice and continuing with Taylor series
$$S_n=2 \left(\sqrt{2}-1\right) \sqrt{n}+\frac{\left(2+\sqrt{2}\right)}{4 \sqrt n} 
   +O\left(\frac{1}{n^{3/2}}\right)$$ Just trying for $n=100$, the value should be $\approx 8.369654$ while the above truncated series gives $\frac{3}{40} \left(267 \sqrt{2}-266\right) \approx 8.369627$
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