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Question-66670

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Question Number 66670 by naka3546 last updated on 18/Aug/19
Commented by mathmax by abdo last updated on 18/Aug/19
let S =Σ_(n=0) ^∞ (3^n /(5^n (n^2 +3n+2))) ⇒S =Σ_(n=0) ^∞ (1/(n^2 +3n+2))((3/5))^n =s((3/5)) with s(x)=Σ_(n=0) ^∞ (x^n /(n^2 +3n+2)) and ∣x∣<1 and x≠0 n^2 +3n+2=0→Δ=9−8=1 ⇒n_1 =((−3+1)/2) =−1 n_2 =((−3−1)/2) =−2 ⇒n^2 +3n+2 =(n+1)(n+2) ⇒ s(x) =Σ_(n=0) ^∞ (x^n /((n+1)(n+2))) =Σ_(n=0) ^∞ ((1/(n+1))−(1/(n+2)))x^n =Σ_(n=0) ^(∞ ) (x^n /(n+1))−Σ_(n=0) ^∞ (x^n /(n+2)) =Σ_(n=1) ^∞ (x^(n−1) /n)−Σ_(n=2) ^∞ (x^(n−2) /n) =(1/x)Σ_(n=1) ^∞ (x^n /n)−(1/x^2 )Σ_(n=2) ^∞ (x^n /n) =(1/x)(−ln(1−x))−(1/x^2 )(−ln(1−x)−1) =−(1/x)ln(1−x)+(1/x^2 )ln(1−x) +(1/x^2 ) ⇒ s((3/5))=−(5/3)ln(1−(3/5))+(9/(25))ln(1−(3/5))+(9/(25)) =−(5/3)ln((2/5))+(9/(25))ln((2/5))+(9/(25)) =((9/(25))−(5/3))ln((2/5))+(9/(25)) =((27−125)/(75))ln((2/5))+(9/(25)) =((98)/(75))ln((2/5))+(9/(25)) (Σ (x^n /n))^′ =Σx^(n−1) =(1/(1−x)) →−ln∣1−x)
$${let}\:{S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\mathrm{5}^{{n}} \left({n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{2}\right)}\:\Rightarrow{S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}} \\ $$$$={s}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)\:\:{with}\:{s}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{2}}\:\:{and}\:\mid{x}\mid<\mathrm{1}\:\:{and}\:{x}\neq\mathrm{0} \\ $$$${n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{2}=\mathrm{0}\rightarrow\Delta=\mathrm{9}−\mathrm{8}=\mathrm{1}\:\Rightarrow{n}_{\mathrm{1}} =\frac{−\mathrm{3}+\mathrm{1}}{\mathrm{2}}\:=−\mathrm{1} \\ $$$${n}_{\mathrm{2}} =\frac{−\mathrm{3}−\mathrm{1}}{\mathrm{2}}\:=−\mathrm{2}\:\Rightarrow{n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{2}\:=\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:\Rightarrow \\ $$$${s}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right){x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty\:\:} \:\frac{{x}^{{n}} }{{n}+\mathrm{1}}−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{n}+\mathrm{2}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}−\mathrm{1}} }{{n}}−\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{{x}^{{n}−\mathrm{2}} }{{n}} \\ $$$$=\frac{\mathrm{1}}{{x}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:=\frac{\mathrm{1}}{{x}}\left(−{ln}\left(\mathrm{1}−{x}\right)\right)−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(−{ln}\left(\mathrm{1}−{x}\right)−\mathrm{1}\right) \\ $$$$=−\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}−{x}\right)\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${s}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)=−\frac{\mathrm{5}}{\mathrm{3}}{ln}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}\right)+\frac{\mathrm{9}}{\mathrm{25}}{ln}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}\right)+\frac{\mathrm{9}}{\mathrm{25}} \\ $$$$=−\frac{\mathrm{5}}{\mathrm{3}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{9}}{\mathrm{25}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{9}}{\mathrm{25}}\:=\left(\frac{\mathrm{9}}{\mathrm{25}}−\frac{\mathrm{5}}{\mathrm{3}}\right){ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{9}}{\mathrm{25}} \\ $$$$=\frac{\mathrm{27}−\mathrm{125}}{\mathrm{75}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{9}}{\mathrm{25}}\:=\frac{\mathrm{98}}{\mathrm{75}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{9}}{\mathrm{25}} \\ $$$$ \\ $$$$\left.\left(\Sigma\:\frac{{x}^{{n}} }{{n}}\right)^{'} \:=\Sigma{x}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\rightarrow−{ln}\mid\mathrm{1}−{x}\right) \\ $$
Answered by mind is power last updated on 18/Aug/19
Σ_(n≥0) (3^n /(5^n (n+1)(n+2)))=Σ_(n≥0) (3^n /5^n )[(1/(n+1))−(1/(n+2))] =Σ_(n≥0) ((((3/5))^n )/(n+1))−Σ_(n≥0) ((((3/5))^n )/(n+2)) ln(1−x)=−Σ_(n≥1) (x^n /n)=−Σ_(n≥0) (x^(n+1) /((n+1))) ∀x∈IR ∣x∣<1 so Σ_(n≥0) ((((3/5))^n )/(n+1))=(5/3)Σ_(m≥0) ((((3/5))^(m+1) )/(m+1))=−(5/3)ln(1−(3/5))=−(5/3)ln((2/5)) −Σ_(n≥0) ((((3/5))^n )/(n+2))=−((25)/9)Σ_(n≥0) ((((3/5))^(n+2) )/(n+2))=((25)/9)[−Σ_(n≥0) ((((3/5))^(n+1) )/(n+1))+(3/5)]=((25)/9)[ln(1−(3/5))+(3/5)] =((25)/9)ln((2/5))+(5/3) Σ_(n≥0) (3^n /(5^n (n^2 +3n+2)))=−(5/3)ln((2/5))+((25)/9)ln((2/5))+(5/3)=−((10)/9)ln((2/5))+(5/3)=−(5/9)[2ln((2/5))−3]
$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}^{{n}} }{\mathrm{5}^{{n}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}^{{n}} }{\mathrm{5}^{{n}} }\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{2}}\right] \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}} }{{n}+\mathrm{1}}−\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}} }{{n}+\mathrm{2}} \\ $$$${ln}\left(\mathrm{1}−{x}\right)=−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{n}} }{{n}}=−\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)}\:\forall{x}\in{IR}\:\mid{x}\mid<\mathrm{1} \\ $$$$ \\ $$$${so}\:\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}} }{{n}+\mathrm{1}}=\frac{\mathrm{5}}{\mathrm{3}}\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}=−\frac{\mathrm{5}}{\mathrm{3}}{ln}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}\right)=−\frac{\mathrm{5}}{\mathrm{3}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right) \\ $$$$−\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}} }{{n}+\mathrm{2}}=−\frac{\mathrm{25}}{\mathrm{9}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}+\mathrm{2}} }{{n}+\mathrm{2}}=\frac{\mathrm{25}}{\mathrm{9}}\left[−\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{5}}\right]=\frac{\mathrm{25}}{\mathrm{9}}\left[{ln}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}\right)+\frac{\mathrm{3}}{\mathrm{5}}\right] \\ $$$$=\frac{\mathrm{25}}{\mathrm{9}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}^{{n}} }{\mathrm{5}^{{n}} \left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right)}=−\frac{\mathrm{5}}{\mathrm{3}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{25}}{\mathrm{9}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{5}}{\mathrm{3}}=−\frac{\mathrm{10}}{\mathrm{9}}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)+\frac{\mathrm{5}}{\mathrm{3}}=−\frac{\mathrm{5}}{\mathrm{9}}\left[\mathrm{2}{ln}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)−\mathrm{3}\right] \\ $$

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