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Question Number 199707 by SANOGO last updated on 08/Nov/23

$$studytheconvergence$$$$\sum ^{}_{n⩾o}sin(\pi \sqrt{4n^2+2}$$

Answered by witcher3 last updated on 09/Nov/23

$$\sqrt{{4\mathrm{n}}^2+2}=2\mathrm{n}\sqrt{1+\frac{1}{{2\mathrm{n}}^2}}$$$$\sin \left(2\pi \mathrm{n}\sqrt{1+\frac{1}{{2\mathrm{n}}^2}}\right)=\sin (2\pi \mathrm{n}\sqrt{1+\frac{1}{{2\mathrm{n}}^2}}-2\pi \mathrm{n})$$$$=\sin (2\pi \mathrm{n}(\sqrt{1+\frac{1}{{2\mathrm{n}}^2}}-1)$$$$\sqrt{1+\frac{1}{{2\mathrm{n}}^2}}-1=\frac{1}{{2\mathrm{n}}^2(\sqrt{1+\frac{1}{{2\mathrm{n}}^2}}+1)}⩽\frac{1}{{4\mathrm{n}}^2}$$$$\sin \left(0\right)⩽\sin (2\pi \mathrm{n}\sqrt{1+\frac{1}{{2\mathrm{n}}^2}}-2\pi \mathrm{n})⩽\sin \left(\frac{\pi }{{2\mathrm{n}}^2}\right)⩽\sin \left(\frac{\pi }{2}\right)$$$$\mathrm{positiv}\mathrm{serie}$$$$2\pi \mathrm{n}\left(\sqrt{1+\frac{1}{{2\mathrm{n}}^2}}\right)-2\pi \mathrm{n}$$$$=2\pi \mathrm{n}(1+\frac{1}{{2\mathrm{n}}^2}-1+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)=\frac{\pi }{\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right)$$$$\sin (\frac{\pi }{\mathrm{n}}+\mathrm{o}\left(\frac{1}{\mathrm{n}}\right))\sim \frac{\pi }{\mathrm{n}};\mathrm{\Sigma }\frac{\pi }{\mathrm{n}}\mathrm{dv}\Rightarrow \mathrm{\Sigma }\sin \left(\pi \sqrt{{4\mathrm{n}}^2+2}\right)\mathrm{dv}$$

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