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Question Number 159379 by HongKing last updated on 16/Nov/21
let S(x) =Σ_(n=0) ^∞ (3x)^(n+2) using the sum above find: Σ_(n=0) ^∞ (((-1)^(n+1) )/(3^(n+1) (n + 3)))
$$\mathrm{let}\:\:\boldsymbol{\mathrm{S}}\left(\mathrm{x}\right)\:=\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{3x}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{2}} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{above}\:\mathrm{find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \left(\mathrm{n}\:+\:\mathrm{3}\right)}\: \\ $$
Answered by Ar Brandon last updated on 16/Nov/21
Σ_(n=0) ^∞ (−(1/3))^(n+1) (1/(n+3))=Σ_(n=0) ^∞ (−(1/3))^(n+1) ∫_0 ^1 x^(n+2) dx =−3Σ_(n=0) ^∞ (−(1/3))^(n+2) ∫_0 ^1 x^(n+2) dx=−3∫_0 ^1 Σ_(n=0) ^∞ (−(x/3))^(n+2) dx =−3∫_0 ^1 ((x^2 /9)/(1+(x/3)))dx=−∫_0 ^1 (x^2 /(3+x))dx=−∫_0 ^1 (((x+3)^2 −6(x+3)+9)/(x+3))dx =−[(x^2 /2)+3x−6x+9ln(x+3)]_0 ^1 =9ln3+(5/2)−9ln4=(5/2)+9ln((3/4))
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}+\mathrm{3}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{2}} {dx} \\ $$$$=−\mathrm{3}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}+\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{2}} {dx}=−\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{{x}}{\mathrm{3}}\right)^{{n}+\mathrm{2}} {dx} \\ $$$$=−\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{{x}^{\mathrm{2}} }{\mathrm{9}}}{\mathrm{1}+\frac{{x}}{\mathrm{3}}}{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{3}+{x}}{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{6}\left({x}+\mathrm{3}\right)+\mathrm{9}}{{x}+\mathrm{3}}{dx} \\ $$$$=−\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}−\mathrm{6}{x}+\mathrm{9ln}\left({x}+\mathrm{3}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{9ln3}+\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{9ln4}=\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{9ln}\left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$
Commented by HongKing last updated on 16/Nov/21
thank you so much my dear Ser cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{cool} \\ $$
Commented by HongKing last updated on 16/Nov/21
my dear Ser, but they said use the first sum S(x) to evaluate the new one
$$\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{but}\:\mathrm{they}\:\mathrm{said}\:\mathrm{use}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{sum}\:\mathrm{S}\left(\mathrm{x}\right)\:\mathrm{to}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{new}\:\mathrm{one} \\ $$
Commented by Ar Brandon last updated on 16/Nov/21
Answered by Ar Brandon last updated on 16/Nov/21
S(x)=Σ_(n=0) ^∞ (3x)^(n+2) =((9x^2 )/(1−3x))=(((1−3x)^2 −2(1−3x)+1)/(1−3x)) ∫S(x)dx=(1/3)Σ_(n=0) ^∞ (((3x)^(n+3) )/(n+3))+C=x−(3/2)x^2 −2x−((ln(1−3x))/3)+C ∫S(0)dx=0=C⇒∫S(x)dx=(1/3)Σ_(n=0) ^∞ (((3x)^(n+3) )/(n+3))=−x−(3/2)x^2 −((ln(1−3x))/3) ∫S(−(1/9))dx=(1/3)Σ_(n=0) ^∞ (((−1)^(n+3) )/(3^(n+3) (n+3)))=(1/9)−(1/(54))−(1/3)ln((4/3)) ⇒Σ_(n=0) ^∞ (((−1)^(n+1) )/(3^(n+1) (n+3)))=3−(1/2)−9ln((4/3))=(5/2)−9ln((4/3))
$${S}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{3}{x}\right)^{{n}+\mathrm{2}} =\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{3}{x}}=\frac{\left(\mathrm{1}−\mathrm{3}{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\mathrm{3}{x}\right)+\mathrm{1}}{\mathrm{1}−\mathrm{3}{x}} \\ $$$$\int{S}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{3}{x}\right)^{{n}+\mathrm{3}} }{{n}+\mathrm{3}}+{C}={x}−\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}{x}−\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{3}{x}\right)}{\mathrm{3}}+{C} \\ $$$$\int{S}\left(\mathrm{0}\right){dx}=\mathrm{0}={C}\Rightarrow\int{S}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{3}{x}\right)^{{n}+\mathrm{3}} }{{n}+\mathrm{3}}=−{x}−\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{3}{x}\right)}{\mathrm{3}} \\ $$$$\int{S}\left(−\frac{\mathrm{1}}{\mathrm{9}}\right){dx}=\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{3}} }{\mathrm{3}^{{n}+\mathrm{3}} \left({n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{54}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{3}^{{n}+\mathrm{1}} \left({n}+\mathrm{3}\right)}=\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{9ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)=\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{9ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$
Commented by HongKing last updated on 16/Nov/21
perfect, thank you so much my dear Ser
$$\mathrm{perfect},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser} \\ $$
Commented by Ar Brandon last updated on 16/Nov/21
You're welcome, Sir.

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