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Question Number 35051 by math khazana by abdo last updated on 14/May/18

calculate Σ_(n=0) ^∞ ((n+3)/(2n+1))x^n

$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}}{x}^{{n}} \\ $$$$ \\ $$

Commented by math khazana by abdo last updated on 10/Jun/18

case1 x>0 let S(x)=Σ_(n=0) ^∞ ((n+3)/(2n+1)) x^n = (1/(√x))Σ_(n=0) ^∞ ((n+3)/(2n+1)) ((√x))^(2n+1) =(1/(√x)) f((√x)) with f(t) = Σ_(n=0) ^∞ ((n+3)/(2n+1)) t^(2n+1) f^′ (t)= Σ_(n=0) ^∞ (n+3)t^(2n) = Σ_(n=0) ^∞ n t^(2n) +3 Σ_(n=0) ^∞ t^(2n) but Σ_(n=0) ^∞ t^(2n) = (1/(1−t^2 )) and we have Σ_(n=0) ^∞ t^n =(1/(1−t)) ⇒ Σ_(n=1) ^∞ n t^(n−1) = (t/((1−t)^2 )) ⇒ Σ_(n=0) ^∞ n t^(2n) = Σ_(n=1) ^∞ n (t^2 )^n =t^2 Σ_(n=1) ^∞ n(t^2 )^(n−1) =t^2 (t^2 /((1−t^2 )^2 )) = (t^4 /((1−t^2 )^2 )) ⇒ f^′ (t) = (t^4 /((1−t^2 )^2 )) +(3/(1−t^2 )) ⇒ f(t) = ∫_. ^t ( (x^4 /((1−x^2 )^2 )) +(3/(1−x^2 )))dx = ∫_. ^t (x^4 /((1−x^2 )^2 ))dx +3 ∫_. ^t (dx/(1−x^2 )) but ∫_. ^t (dx/(1−x^2 )) = ∫^t { (1/(1−x)) +(1/(1+x))}dx=ln∣((1+t)/(1−t))∣ +c changement x =chu give ∫^t (x^4 /((1−x^2 )^2 ))dx = ∫^t ((ch^4 u)/(sh^4 u)) .shu du =∫^t ((ch^4 u)/(sh^3 u)) du = (1/4)∫^t (((1+ch(2u))^2 )/(((ch(2u)−1)/2) sh(u))) du =(1/2) ∫^t ((1+2ch(2u) +ch^2 (2u))/((ch(2u)−1)sh(u))) du ....be continued..

$${case}\mathrm{1}\:\:{x}>\mathrm{0}\:\:{let}\:\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}}\:{x}^{{n}} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{{x}}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}}\:\left(\sqrt{{x}}\right)^{\mathrm{2}{n}+\mathrm{1}} \:=\frac{\mathrm{1}}{\sqrt{{x}}}\:{f}\left(\sqrt{{x}}\right)\:{with} \\ $$$${f}\left({t}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}+\mathrm{3}}{\mathrm{2}{n}+\mathrm{1}}\:{t}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${f}^{'} \left({t}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left({n}+\mathrm{3}\right){t}^{\mathrm{2}{n}} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{n}\:{t}^{\mathrm{2}{n}} \:\:+\mathrm{3}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}{n}} \:\:\:{but} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{t}^{\mathrm{2}{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\:{and}\:{we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{t}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\:{t}^{{n}−\mathrm{1}} \:=\:\frac{{t}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{n}\:{t}^{\mathrm{2}{n}} \:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\:\left({t}^{\mathrm{2}} \right)^{{n}} \:={t}^{\mathrm{2}} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\left({t}^{\mathrm{2}} \right)^{{n}−\mathrm{1}} \\ $$$$={t}^{\mathrm{2}} \:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\frac{{t}^{\mathrm{4}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)\:=\:\frac{{t}^{\mathrm{4}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\:\int_{.} ^{{t}} \:\:\left(\:\:\frac{{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\:\int_{.} ^{{t}} \:\:\:\:\frac{{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:+\mathrm{3}\:\int_{.} ^{{t}} \:\:\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\:{but} \\ $$$$\int_{.} ^{{t}} \:\:\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} }\:=\:\int^{{t}} \:\:\left\{\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:+\frac{\mathrm{1}}{\mathrm{1}+{x}}\right\}{dx}={ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:+{c} \\ $$$${changement}\:{x}\:={chu}\:{give} \\ $$$$\int^{{t}} \:\:\:\:\frac{{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\:\int^{{t}} \:\:\frac{{ch}^{\mathrm{4}} {u}}{{sh}^{\mathrm{4}} {u}}\:.{shu}\:{du} \\ $$$$=\int^{{t}} \:\:\:\:\frac{{ch}^{\mathrm{4}} {u}}{{sh}^{\mathrm{3}} {u}}\:{du}\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int^{{t}} \:\:\:\:\frac{\left(\mathrm{1}+{ch}\left(\mathrm{2}{u}\right)\right)^{\mathrm{2}} }{\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}\:{sh}\left({u}\right)}\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int^{{t}} \:\:\:\frac{\mathrm{1}+\mathrm{2}{ch}\left(\mathrm{2}{u}\right)\:+{ch}^{\mathrm{2}} \left(\mathrm{2}{u}\right)}{\left({ch}\left(\mathrm{2}{u}\right)−\mathrm{1}\right){sh}\left({u}\right)}\:{du}\:....{be}\:{continued}.. \\ $$$$ \\ $$

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