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Question Number 65324 by hovea cw last updated on 28/Jul/19

Commented by mathmax by abdo last updated on 28/Jul/19

$$wehave1⩽k⩽n^2\Rightarrow 1+n^2⩽k+n^2⩽2n^2\Rightarrow \sqrt{1+n^2}⩽\sqrt{k+n^2}⩽n\sqrt{2}\Rightarrow$$$$\frac{1}{n\sqrt{2}}⩽\frac{1}{\sqrt{k+n^2}}⩽\frac{1}{\sqrt{1+n^2}}\Rightarrow \sum _{k=1}^{n^2}\frac{1}{n\sqrt{2}}⩽\sum _{k=1}^{n^2}\frac{1}{\sqrt{k+n^2}}⩽\sum _{k=1}^{n^2}\frac{1}{\sqrt{1+n^2}}$$$$\Rightarrow \frac{n^2}{n\sqrt{2}}⩽\sum _{k=1}^{n^2}\frac{1}{\sqrt{k+n^2}}\Rightarrow S_n⩾\frac{n}{\sqrt{2}}but{lim}_{n\to +\mathrm{\infty }}\frac{n}{\sqrt{2}}=+\mathrm{\infty }\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=+\mathrm{\infty }$$

Commented by Cmr 237 last updated on 27/Aug/19

$$\mathrm{waouh}\mathrm{chapeau}$$

Commented by mathmax by abdo last updated on 27/Aug/19

$$thanks.$$

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