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Question Number 52670 by maxmathsup by imad last updated on 11/Jan/19

study the convergence of Σ_(n=0) ^∞ sin(π(√(4n^2 +1)))

$${study}\:{the}\:{convergence}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$

Commented by maxmathsup by imad last updated on 12/Jan/19

we have π(√(4n^2 +1))=π(√(4n^2 (1+(1/(4n^2 ))))) =2nπ(√(1+(1/(4n^2 )))) but we have (√(1+x))=(1+x)^(1/2) =1+(x/2) +(((1/2)((1/(2 ))−1))/2) x^2 +o(x^3 ) (x→0) ⇒ =1+(x/2) −(1/8) x^2 +o(x^3 ) ⇒(√(1+(1/(4n^2 ))))=1+(1/(8n^2 )) −(1/(8.16n^4 )) +o((1/n^6 )) ⇒ 2nπ(√(1+(1/(4n^2 ))))=2nπ +(π/(4n)) −(π/(32n^3 )) +o((1/n^5 ))⇒sin(π(√(4n^2 +1))) ∼ sin((π/(4n)))∼(π/(4n)) but Σ (π/(4n)) diverges ⇒Σ sin(π(√(4n^2 +1))) diverges.

$$\left.{we}\:{have}\:\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}=\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }\right.}\right)\:\:=\mathrm{2}{n}\pi\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }}\:\:\:{but}\:{we}\:{have} \\ $$$$\sqrt{\mathrm{1}+{x}}=\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\mathrm{1}+\frac{{x}}{\mathrm{2}}\:+\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}\:}−\mathrm{1}\right)}{\mathrm{2}}\:{x}^{\mathrm{2}} \:+{o}\left({x}^{\mathrm{3}} \right)\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$=\mathrm{1}+\frac{{x}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{8}}\:{x}^{\mathrm{2}} \:+{o}\left({x}^{\mathrm{3}} \right)\:\Rightarrow\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{8}.\mathrm{16}{n}^{\mathrm{4}} }\:+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{6}} }\right)\:\Rightarrow \\ $$$$\mathrm{2}{n}\pi\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }}=\mathrm{2}{n}\pi\:+\frac{\pi}{\mathrm{4}{n}}\:−\frac{\pi}{\mathrm{32}{n}^{\mathrm{3}} }\:+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{5}} }\right)\Rightarrow{sin}\left(\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}\right)\:\sim\:{sin}\left(\frac{\pi}{\mathrm{4}{n}}\right)\sim\frac{\pi}{\mathrm{4}{n}} \\ $$$${but}\:\Sigma\:\frac{\pi}{\mathrm{4}{n}}\:{diverges}\:\Rightarrow\Sigma\:{sin}\left(\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} \:+\mathrm{1}}\right)\:{diverges}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jan/19

((4n^2 +1)/2)≥(√(4n^2 ×1)) 4n^2 +1≥4n (√(4n^2 +1)) ≥(√(4n)) sin(π(√(4n^2 +1)) )≈sin(π×2(√n) ) sin((√n) ×2π) 1)now when n is perfect square sin((√n) ×2π)=0 2)(√n) =I+f I=integer f=fractional part sin{(I+f)2π} =sin{I×2π+f×2π} =sin(f×2π) 1≥sin(2πf)≥−1 so Σ_(n=0) ^∞ sin(π(√(4n^2 +1)) ) does not converge Σ_(n=0) ^∞ sin(π(√(4n^2 +1)) ) oscilates between ± ∞ i have tried to solve ...pls check is the logic true...

$$\frac{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}\geqslant\sqrt{\mathrm{4}{n}^{\mathrm{2}} ×\mathrm{1}}\: \\ $$$$\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{4}{n} \\ $$$$\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}\:\geqslant\sqrt{\mathrm{4}{n}}\: \\ $$$${sin}\left(\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}\:\right)\approx{sin}\left(\pi×\mathrm{2}\sqrt{{n}}\:\right) \\ $$$${sin}\left(\sqrt{{n}}\:×\mathrm{2}\pi\right) \\ $$$$\left.\mathrm{1}\right){now}\:{when}\:{n}\:{is}\:{perfect}\:{square} \\ $$$${sin}\left(\sqrt{{n}}\:×\mathrm{2}\pi\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\sqrt{{n}}\:={I}+{f}\:{I}={integer} \\ $$$${f}={fractional}\:{part} \\ $$$$ \\ $$$${sin}\left\{\left({I}+{f}\right)\mathrm{2}\pi\right\} \\ $$$$={sin}\left\{{I}×\mathrm{2}\pi+{f}×\mathrm{2}\pi\right\} \\ $$$$={sin}\left({f}×\mathrm{2}\pi\right) \\ $$$$\:\mathrm{1}\geqslant{sin}\left(\mathrm{2}\pi{f}\right)\geqslant−\mathrm{1} \\ $$$${so}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{sin}\left(\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}\:\right)\:{does}\:{not}\:{converge} \\ $$$$\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{sin}\left(\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{1}}\:\right)\:{oscilates}\:{between}\:\pm\:\infty \\ $$$${i}\:{have}\:{tried}\:{to}\:{solve}\:...{pls}\:{check}\:{is}\:{the}\:{logic} \\ $$$${true}... \\ $$

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