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Question Number 51185 by Abdo msup. last updated on 24/Dec/18
calculate Σ_(n=0) ^∞ (n/((n+1)^4 (2n+1)^2 ))
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by Abdo msup. last updated on 30/Dec/18
let decompose F(x)=(x/((x+1)^4 (2x+1)^2 )) F(x)=(a/(x+1)) +(b/((x+1)^2 )) +(c/((x+1)^3 )) +(d/((x+1)^4 )) +(e/(2x+1)) +(f/((2x+1)^2 )) d =lim_(x→−1) (x+1)^4 F(x)=−1 f =lim_(x→−(1/2)) (2x+1)^2 F(x)=((−1)/2).16 =−8 ⇒ F(x)=(a/(x+1)) +(b/((x+1)^2 )) +(c/((x+1)^3 )) −(1/((x+1)^4 )) +(e/(2x+1)) −(8/((2x+1)^2 )) lim_(x→+∞) xF(x)=0 =a+(e/2) ⇒2a+e=0 ⇒e=−2a ⇒F(x)=(a/(x+1)) +(b/((x+1)^2 )) +(c/((x+1)^3 )) −(1/((x+1)^4 )) −((2a)/(2x+1)) −(8/((2x+1)^2 )) F(0)=0=a+b+c−1−2a−8=−a+b+c−9 ⇒ −a+b+c =9 ⇒a−b−c=−9 F(−2)=((−2)/9) =−a+b−c−1+(2/3)a−(8/9) ⇒ −2 =−9a+9b−9c−9+6a−8 ⇒ −3a+9b−9c−15=0 ⇒3a−9b+9c+15=0 ⇒ a−3b+3c=−5...be continued...
$${let}\:{decompose}\:{F}\left({x}\right)=\frac{{x}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$+\frac{{e}}{\mathrm{2}{x}+\mathrm{1}}\:+\frac{{f}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${d}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{4}} {F}\left({x}\right)=−\mathrm{1} \\ $$$${f}\:={lim}_{{x}\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\frac{−\mathrm{1}}{\mathrm{2}}.\mathrm{16}\:=−\mathrm{8}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$+\frac{{e}}{\mathrm{2}{x}+\mathrm{1}}\:−\frac{\mathrm{8}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+\frac{{e}}{\mathrm{2}}\:\Rightarrow\mathrm{2}{a}+{e}=\mathrm{0}\:\Rightarrow{e}=−\mathrm{2}{a} \\ $$$$\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{b}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$−\frac{\mathrm{2}{a}}{\mathrm{2}{x}+\mathrm{1}}\:−\frac{\mathrm{8}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}={a}+{b}+{c}−\mathrm{1}−\mathrm{2}{a}−\mathrm{8}=−{a}+{b}+{c}−\mathrm{9}\:\Rightarrow \\ $$$$−{a}+{b}+{c}\:=\mathrm{9}\:\Rightarrow{a}−{b}−{c}=−\mathrm{9} \\ $$$${F}\left(−\mathrm{2}\right)=\frac{−\mathrm{2}}{\mathrm{9}}\:=−{a}+{b}−{c}−\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}{a}−\frac{\mathrm{8}}{\mathrm{9}}\:\Rightarrow \\ $$$$−\mathrm{2}\:=−\mathrm{9}{a}+\mathrm{9}{b}−\mathrm{9}{c}−\mathrm{9}+\mathrm{6}{a}−\mathrm{8}\:\Rightarrow \\ $$$$−\mathrm{3}{a}+\mathrm{9}{b}−\mathrm{9}{c}−\mathrm{15}=\mathrm{0}\:\Rightarrow\mathrm{3}{a}−\mathrm{9}{b}+\mathrm{9}{c}+\mathrm{15}=\mathrm{0}\:\Rightarrow \\ $$$${a}−\mathrm{3}{b}+\mathrm{3}{c}=−\mathrm{5}…{be}\:{continued}… \\ $$

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