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Question Number 29981 by abdo imad last updated on 14/Feb/18
find radius and sum of Σ_(n=0) ^∞ (x^(2n) /(2n+1)) 2) find Σ_(n=0) ^∞ (1/((2n+1)9^n )) .
$${find}\:{radius}\:{and}\:{sum}\:{of}\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{9}^{{n}} }\:. \\ $$
Commented by abdo imad last updated on 15/Feb/18
let put for ∣x∣<1S(x)= Σ_(n=0) ^∞ (x^(2n) /(2n+1))⇒x S(x)=Σ_(n=0) ^∞ (x^(2n+1) /(2n+1)) let derivate S(x)+xS^′ (x)= Σ_(n=0) ^∞ x^(2n) = (1/(1−x^2 )) ⇒ S is solution of d.e xy^′ +y = (1/(1−x^2 )) h.e⇒ xy^′ +y =0 ⇒ (y^′ /y) =−(1/x) ⇒ ln∣y∣=−lnx +c ⇒y= (k/x) let use m.v.c we have y^′ =((k^′ x−k)/x^2 ) (e)⇒((k^′ x −k)/x) + (k/x)= (1/(1−x^2 ))⇒k^′ = (1/(1−x^2 )) k(x)= ∫(dx/(1−x^2 )) +λ =(1/2)( ∫ (dx/(1−x)) +∫ (dx/(1+x))) +λ =(1/2)ln∣((1+x)/(1−x))∣+λ ⇒S(x) = (1/(2x))ln∣((1+x)/(1−x))∣ +(λ/x) λ=lim_(x→0) (xS(x)−(1/2)ln∣((1+x)/(1−x))∣) =0 ⇒ S(x)=(1/(2x))ln∣((1+x)/(1−x))∣ with −1<x<1 . we have (u_(n+1) /u_n )=((1/(2n+3))/(1/(2n+1)))= ((2n+1)/(2n+3_(n→+∞) )) →1 ⇒R=1 for x=1 or x=−1 the serie diverges. 2)Σ_(n=0) ^∞ (1/((2n+1)9^n ))=Σ_(n=0) ^∞ ((((1/3))^(2n) )/(2n+1)) =S((1/3)) = (3/2)ln∣((4/3)/(2/3))∣=(3/2)ln(2) .
$${let}\:{put}\:\:{for}\:\mid{x}\mid<\mathrm{1}{S}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}\Rightarrow{x}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:{let}\:{derivate} \\ $$$${S}\left({x}\right)+{xS}^{'} \left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{\mathrm{2}{n}} =\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow\:{S}\:{is}\:{solution}\:{of}\:{d}.{e} \\ $$$${xy}^{'} \:+{y}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:{h}.{e}\Rightarrow\:{xy}^{'} \:+{y}\:=\mathrm{0}\:\Rightarrow\:\frac{{y}^{'} }{{y}}\:=−\frac{\mathrm{1}}{{x}}\:\Rightarrow \\ $$$${ln}\mid{y}\mid=−{lnx}\:+{c}\:\Rightarrow{y}=\:\frac{{k}}{{x}}\:\:\:{let}\:{use}\:{m}.{v}.{c}\:{we}\:{have} \\ $$$${y}^{'} =\frac{{k}^{'} {x}−{k}}{{x}^{\mathrm{2}} }\:\:\:\left({e}\right)\Rightarrow\frac{{k}^{'} {x}\:−{k}}{{x}}\:+\:\frac{{k}}{{x}}=\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\Rightarrow{k}^{'} =\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${k}\left({x}\right)=\:\int\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} }\:+\lambda\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\int\:\frac{{dx}}{\mathrm{1}−{x}}\:+\int\:\frac{{dx}}{\mathrm{1}+{x}}\right)\:+\lambda \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid+\lambda\:\Rightarrow{S}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}{x}}{ln}\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\:+\frac{\lambda}{{x}} \\ $$$$\lambda={lim}_{{x}\rightarrow\mathrm{0}} \left({xS}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\right)\:=\mathrm{0}\:\Rightarrow \\ $$$${S}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{x}}{ln}\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\:\:{with}\:\:−\mathrm{1}<{x}<\mathrm{1}\:\:. \\ $$$${we}\:{have}\:\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }=\frac{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}}{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}}=\:\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}_{{n}\rightarrow+\infty} }\:\rightarrow\mathrm{1}\:\Rightarrow{R}=\mathrm{1} \\ $$$${for}\:{x}=\mathrm{1}\:{or}\:{x}=−\mathrm{1}\:{the}\:{serie}\:{diverges}. \\ $$$$\left.\mathrm{2}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{9}^{{n}} }=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}\:={S}\left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}{ln}\mid\frac{\frac{\mathrm{4}}{\mathrm{3}}}{\frac{\mathrm{2}}{\mathrm{3}}}\mid=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:. \\ $$

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