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Question Number 165493 by SANOGO last updated on 02/Feb/22

calcul la somme de cette serie entiere Σ_(n=0) ^(+oo) ((2n+3)/(2n+1))x^n

$${calcul}\:{la}\:{somme}\:{de}\:{cette}\:{serie}\:{entiere} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{+{oo}} {\sum}}\frac{\mathrm{2}{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}}{x}^{{n}} \\ $$

Answered by TheSupreme last updated on 02/Feb/22

S=Σ_(n=0) ^∞ (1+(2/(2n+1)))x^n S=(1/(1−x))+Σ(2/(2n+1))x^n x=s^2 S=(1/(1−s^2 ))+(2/s)Σ(1/(2n+1))s^(2n+1) Σ(1/(2n+1))s^(2n+1) =Σ∫s^(2n) =∫Σs^(2n) =∫(1/(1−s^2 ))=arctanh(s) S=(1/(1−x))+(1/( (√x)))arctanh((√x)) ∣x∣<1

$${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\right){x}^{{n}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{1}−{x}}+\Sigma\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}{x}^{{n}} \\ $$$${x}={s}^{\mathrm{2}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{1}−{s}^{\mathrm{2}} }+\frac{\mathrm{2}}{{s}}\Sigma\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{s}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{s}^{\mathrm{2}{n}+\mathrm{1}} =\Sigma\int{s}^{\mathrm{2}{n}} =\int\Sigma{s}^{\mathrm{2}{n}} =\int\frac{\mathrm{1}}{\mathrm{1}−{s}^{\mathrm{2}} }={arctanh}\left({s}\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}{arctanh}\left(\sqrt{{x}}\right) \\ $$$$\mid{x}\mid<\mathrm{1} \\ $$

Answered by Mathspace last updated on 02/Feb/22

S=Σ_(n=0) ^∞ x^n +2Σ_(n=0) ^∞ (x^n /(2n+1)) for ∣x∣<1 Σ_(n=0) ^∞ x^n =(1/(1−x)) Σ_(n=0) ^∞ (x^n /(2n+1))=(1/( (√x)))Σ_(n=0) ^(∞ ) ((((√x))^(2n+1) )/(2n+1)) =(1/( (√x)))ϕ((√x)) with ϕ(t)=Σ_(n=0) ^(∞ ) (t^(2n+1) /(2n+1)) ϕ^′ (t)=Σ_(n=0) ^∞ t^(2n) =(1/(1−t^2 )) ⇒ ϕ(t)=∫(dt/(1−t^2 )) +C =(1/2)∫((1/(1−t))+(1/(1+t)))dt =(1/2)ln∣((1+t)/(1−t))∣ ⇒ϕ(t)=(1/2)ln∣((1+t)/(1−t))∣ +C ϕ(0)=0=C ⇒ϕ(t)=(1/2)ln∣((1+t)/(1−t))∣ ⇒ S=(1/(1−x)) +(2/( (√x)))ϕ((√x)) =(1/(1−x))+(1/( (√x)))ln∣((1+(√x))/(1−(√x)))∣

$${S}=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} +\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$${for}\:\mid{x}\mid<\mathrm{1}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{{x}}}\sum_{{n}=\mathrm{0}} ^{\infty\:} \:\frac{\left(\sqrt{{x}}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{x}}}\varphi\left(\sqrt{{x}}\right)\:{with}\:\varphi\left({t}\right)=\sum_{{n}=\mathrm{0}} ^{\infty\:} \frac{{t}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\varphi^{'} \left({t}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\varphi\left({t}\right)=\int\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:\Rightarrow\varphi\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:+{C} \\ $$$$\varphi\left(\mathrm{0}\right)=\mathrm{0}={C}\:\Rightarrow\varphi\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:\Rightarrow \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:+\frac{\mathrm{2}}{\:\sqrt{{x}}}\varphi\left(\sqrt{{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}{ln}\mid\frac{\mathrm{1}+\sqrt{{x}}}{\mathrm{1}−\sqrt{{x}}}\mid \\ $$

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