Show that $$\lim{\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}+...\frac{1}{\sqrt{2n}}}=\infty$$.
not allowed to use series test and any result/test of series. Only things allowed are use of first principles $\epsilon $ $\delta$ definition , cauchy's convergence etc.
I tried to use $Cauchy's$ $ Criterion$ for convergence of sequences since $S_{n+1}-S_{n}$ comes out nicely . But it doesn't leads to result. Hints are appreciated .
Have a good day !
We know that $$\lim_{n \to \infty} \frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + ... + \frac{1}{\sqrt{2n}} \geq \lim_{n \to \infty} \frac{1}{\sqrt{2n}} + \frac{1}{\sqrt{2n}} + ... + \frac{1}{\sqrt{2n}} $$ And $$ \lim_{n \to \infty} \frac{1}{\sqrt{2n}} + \frac{1}{\sqrt{2n}} + ... + \frac{1}{\sqrt{2n}} = \lim_{n \to \infty} \frac{n}{\sqrt{2n}} = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{2}} = \infty $$
So
$$\therefore \lim_{n \to \infty} \frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + ... + \frac{1}{\sqrt{2n}} = \infty$$