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Question Number 166511 by mkam last updated on 21/Feb/22
is the function f(x) = tan^(−1) (x) + tan^(−1) (2x) + tan^(−1) (3x) + .....   converge or diverge ?
$${is}\:{the}\:{function}\:{f}\left({x}\right)\:=\:{tan}^{−\mathrm{1}} \left({x}\right)\:+\:{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\:+\:{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}\right)\:+\:…..\: \\ $$$${converge}\:{or}\:{diverge}\:? \\ $$$$ \\ $$
Answered by alephzero last updated on 21/Feb/22
f(x) = Σ_(n=1) ^∞ arctg nx converges, if  lim_(n→∞) ((arctg (n+1)x)/(arctg nx)) < 1  1. f(x) converges at 0.  f(0) = arctg 0 + arctg 0 + ... = 0  2.f(x) diverges, if x ≠ 0  f(x) = Σ_(n=1) ^∞  arctg nx  2.1 ρ = lim_(n→∞) ((arctg (n+1)x)/(arctg nx))  (there x ≠ 0)  If ρ < 1 ⇒ Σ_(n=1) ^∞  arctg nx converges.  If ρ > 1 ⇒ Σ_(n=1) ^∞  arctg nx diverges.  If ρ = 1 ⇒ Σ_(n=1) ^∞  arctg nx can  converge and diverge.  ρ = lim_(n→∞) ((arctg (n+1)x)/(arctg nx)) =  = ((lim_(n→∞)  arctg (n+1)x)/(lim_(n→∞)  arctg nx)) = ((π/2)/(π/2)) = 1  2.2 σ = lim_(n→∞) ((arctg nx))^(1/n)   If σ < 1 ⇒ f(x) converges  If σ > 1 ⇒ f(x) diverges  If σ = 1 ⇒ f(x) can converge and  diverge.  σ = lim_(n→∞) ((arctg nx))^(1/n)  = 1  I think f(x) diverges, if x ≠ 0.
$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctg}\:{nx}\:\mathrm{converges},\:\mathrm{if} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{arctg}\:\left({n}+\mathrm{1}\right){x}}{\mathrm{arctg}\:{nx}}\:<\:\mathrm{1} \\ $$$$\mathrm{1}.\:{f}\left({x}\right)\:\mathrm{converges}\:\mathrm{at}\:\mathrm{0}. \\ $$$${f}\left(\mathrm{0}\right)\:=\:\mathrm{arctg}\:\mathrm{0}\:+\:\mathrm{arctg}\:\mathrm{0}\:+\:…\:=\:\mathrm{0} \\ $$$$\mathrm{2}.{f}\left({x}\right)\:\mathrm{diverges},\:\mathrm{if}\:{x}\:\neq\:\mathrm{0} \\ $$$${f}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{arctg}\:{nx} \\ $$$$\mathrm{2}.\mathrm{1}\:\rho\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{arctg}\:\left({n}+\mathrm{1}\right){x}}{\mathrm{arctg}\:{nx}} \\ $$$$\left(\mathrm{there}\:{x}\:\neq\:\mathrm{0}\right) \\ $$$$\mathrm{If}\:\rho\:<\:\mathrm{1}\:\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{arctg}\:{nx}\:\mathrm{converges}. \\ $$$$\mathrm{If}\:\rho\:>\:\mathrm{1}\:\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{arctg}\:{nx}\:\mathrm{diverges}. \\ $$$$\mathrm{If}\:\rho\:=\:\mathrm{1}\:\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{arctg}\:{nx}\:\mathrm{can} \\ $$$$\mathrm{converge}\:\mathrm{and}\:\mathrm{diverge}. \\ $$$$\rho\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{arctg}\:\left({n}+\mathrm{1}\right){x}}{\mathrm{arctg}\:{nx}}\:= \\ $$$$=\:\frac{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{arctg}\:\left({n}+\mathrm{1}\right){x}}{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{arctg}\:{nx}}\:=\:\frac{\frac{\pi}{\mathrm{2}}}{\frac{\pi}{\mathrm{2}}}\:=\:\mathrm{1} \\ $$$$\mathrm{2}.\mathrm{2}\:\sigma\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{{n}}]{\mathrm{arctg}\:{nx}} \\ $$$$\mathrm{If}\:\sigma\:<\:\mathrm{1}\:\Rightarrow\:{f}\left({x}\right)\:\mathrm{converges} \\ $$$$\mathrm{If}\:\sigma\:>\:\mathrm{1}\:\Rightarrow\:{f}\left({x}\right)\:\mathrm{diverges} \\ $$$$\mathrm{If}\:\sigma\:=\:\mathrm{1}\:\Rightarrow\:{f}\left({x}\right)\:\mathrm{can}\:\mathrm{converge}\:\mathrm{and} \\ $$$$\mathrm{diverge}. \\ $$$$\sigma\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{{n}}]{\mathrm{arctg}\:{nx}}\:=\:\mathrm{1} \\ $$$$\mathrm{I}\:\mathrm{think}\:{f}\left({x}\right)\:\mathrm{diverges},\:\mathrm{if}\:{x}\:\neq\:\mathrm{0}. \\ $$

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