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Question Number 166511 by mkam last updated on 21/Feb/22

$$isthefunctionf\left(x\right)={tan}^{-1}\left(x\right)+{tan}^{-1}\left(2x\right)+{tan}^{-1}\left(3x\right)+.....$$$$convergeordiverge?$$$$$$

Answered by alephzero last updated on 21/Feb/22

$$f\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{arctg}nx\mathrm{converges},\mathrm{if}$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{\mathrm{arctg}(n+1)x}{\mathrm{arctg}nx}<1$$$$1.f\left(x\right)\mathrm{converges}\mathrm{at}0.$$$$f\left(0\right)=\mathrm{arctg}0+\mathrm{arctg}0+...=0$$$$2.f\left(x\right)\mathrm{diverges},\mathrm{if}x\ne 0$$$$f\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{arctg}nx$$$$2.1\rho =\underset{n\to \mathrm{\infty }}{\lim }\frac{\mathrm{arctg}(n+1)x}{\mathrm{arctg}nx}$$$$(\mathrm{there}x\ne 0)$$$$\mathrm{If}\rho <1\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{arctg}nx\mathrm{converges}.$$$$\mathrm{If}\rho >1\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{arctg}nx\mathrm{diverges}.$$$$\mathrm{If}\rho =1\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{arctg}nx\mathrm{can}$$$$\mathrm{converge}\mathrm{and}\mathrm{diverge}.$$$$\rho =\underset{n\to \mathrm{\infty }}{\lim }\frac{\mathrm{arctg}(n+1)x}{\mathrm{arctg}nx}=$$$$=\frac{\underset{n\to \mathrm{\infty }}{\lim }\mathrm{arctg}(n+1)x}{\underset{n\to \mathrm{\infty }}{\lim }\mathrm{arctg}nx}=\frac{\frac{\pi }{2}}{\frac{\pi }{2}}=1$$$$2.2\sigma =\underset{n\to \mathrm{\infty }}{\lim }\sqrt[n]{\mathrm{arctg}nx}$$$$\mathrm{If}\sigma <1\Rightarrow f\left(x\right)\mathrm{converges}$$$$\mathrm{If}\sigma >1\Rightarrow f\left(x\right)\mathrm{diverges}$$$$\mathrm{If}\sigma =1\Rightarrow f\left(x\right)\mathrm{can}\mathrm{converge}\mathrm{and}$$$$\mathrm{diverge}.$$$$\sigma =\underset{n\to \mathrm{\infty }}{\lim }\sqrt[n]{\mathrm{arctg}nx}=1$$$$\mathrm{I}\mathrm{think}f\left(x\right)\mathrm{diverges},\mathrm{if}x\ne 0.$$

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