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Question Number 203774 by Calculusboy last updated on 27/Jan/24

$$\mathit{d}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{n}\mathit{e}\mathit{w}\mathit{h}\mathit{e}\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}\mathit{i}\mathit{s}$$$$\mathit{c}\mathit{o}\mathit{n}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{n}\mathit{t}\mathit{o}\mathit{r}\mathit{d}\mathit{i}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{n}\mathit{t}$$$$\underset{\mathit{n}=1}{\mathbf{\sum }^{\mathrm{\infty }}}\frac{\mathit{n}}{\sqrt{4{\mathit{n}}^2+1}}$$

Answered by witcher3 last updated on 27/Jan/24

$$\frac{\mathrm{n}}{\sqrt{{4\mathrm{n}}^2+1}}⩾\frac{\mathrm{n}}{\sqrt{1+{4\mathrm{n}}^2+4\mathrm{n}}}=\frac{\mathrm{n}}{2\mathrm{n}+1}=\frac{3\mathrm{n}}{3(2\mathrm{n}+1)}>\frac{2\mathrm{n}+1}{3(2\mathrm{n}+1)}$$$$\Rightarrow \mathrm{\Sigma }\frac{\mathrm{n}}{\sqrt{{4\mathrm{n}}^2+1}}>\mathrm{\Sigma }\frac{1}{3}\mathrm{Dv}$$$$$$

Answered by Mathspace last updated on 30/Jan/24

$$\frac{n}{\sqrt{4n^2+1}}=\frac{n}{2n\sqrt{1+\frac{1}{4n^2}}}=\frac{1}{2\sqrt{1+\frac{1}{4n^2}}}$$$$wehave{(1+x)}^{\alpha }=1+\alpha x+\frac{\alpha (\alpha -1)}{2}{x}^2+o\left(x^2\right)\Rightarrow$$$$\frac{1}{\sqrt{1+x}}={(1+x)}^{-\frac{1}{2}}$$$$=1-\frac{x}{2}+\frac{1}{2}(-\frac{1}{2})(-\frac{1}{2}-1)x^2+o\left(x^2\right)$$$$=1-\frac{x}{2}-\frac{1}{4}\times \frac{-3}{2}{x}^2+o\left(x^2\right)$$$$=1-\frac{x}{2}+\frac{3}{8}{x}^2+o\left(x^2\right)so$$$$u_n=\frac{1}{2\sqrt{1+\frac{1}{4n^2}}}=\frac{1}{2}-\frac{1}{16n^2}+\frac{3}{16}\times \frac{1}{16n^4}+o\left(\frac{1}{n^4}\right)$$$$estundev.assymtotiquede$$$$u_{n}onalim{u}_n=\frac{1}{2}\ne 0\Rightarrow$$$$\mathrm{\Sigma }{u}_{n}estdivergente.$$$$$$

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