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Question Number 69786 by mathmax by abdo last updated on 27/Sep/19

calculate Σ_(n=1) ^∞ (((2n+1)(−1)^n )/(n^2 (n+1)(n+2)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$

Commented by mathmax by abdo last updated on 02/Oct/19

let S_n =Σ_(k=1) ^n (((2k+1))/(k^2 (k+1)(k+2)^2 ))(−1)^k we have S=lim_(n→+∞) S_n let decompose F(x)=((2x+1)/(x^2 (x+1)(x+2)^2 )) F(x)=(a/x)+(b/x^2 ) +(c/(x+1)) +(d/(x+2)) +(e/((x+2)^2 )) b =lim_(x→0) x^2 F(x)=(1/4) e =lim_(x→−2) (x+2)^2 F(x) =((−3)/(−4))=(3/4) ⇒ F(x)=(a/x)+(1/(4x^2 )) +(c/(x+1)) +(d/(x+2)) +(3/(4(x+2)^2 )) lim_(x→+∞) xF(x)=0 =a+c+d F(1)=(3/(18)) =(1/6) =a+(1/4) +(c/2) +(d/3) +(1/(12)) =a+(c/2) +(d/3) +(1/3) ⇒ 1 =6a+3c+2d+2 ⇒6a+3c+2d=−1 F(−3) =((−5)/(9(−2))) =(5/(18)) =−(a/3)+(1/(36))−(c/2)−d+(3/4) =−(a/3)−(c/2)−d +(7/9) ⇒5=−6a−9c−18d+14 ⇒ −6a−9c−18d =−9 ⇒6a+9c+18d =9 ⇒2a+3c+6d =3 we get the system { ((a+c+d=0)),((6a+3c+2d =−1)) :} {2a+3c+6d =3 ⇒d=−a−c and { ((6a+3c+2(−a−c)=−1)),((2a+3c+6(−a−c)=3 ⇒ { ((4a+c=−1)),((−4a−3c =3 ⇒−2c=2 ⇒c=−1)) :})) :} 4a=−1+1 =0 ⇒a=0 ⇒d=1 ⇒ F(x)=(1/(4x^2 ))−(1/(x+1)) +(1/(x+2)) +(3/(4(x+2)^2 )) ⇒ S_n =Σ_(k=1) ^n (−1)^k F(k) =(1/4)Σ_(k=1) ^n (((−1)^k )/k^2 )−Σ_(k=1) ^n (((−1)^k )/(k+1)) +Σ_(k=1) ^n (((−1)^k )/(k+2)) +(3/4)Σ_(k=1) ^n (((−1)^k )/((k+2)^2 )) and lim_(n→+∞) S_n =(1/4)Σ_(k=1) ^∞ (((−1)^k )/k^2 )−Σ_(k=2) ^∞ (((−1)^(k−1) )/k) +Σ_(k=3) ^∞ (((−1)^(k−2) )/k)+(3/4)Σ_(k=3) ^∞ (((−1)^(k−2) )/k^2 ) Σ_(k=1) ^∞ (((−1)^k )/k^2 ) =(2^(1−2) −1)ξ(2) =−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(k=2) ^∞ (((−1)^(k−1) )/k) =−Σ_(k=2) ^∞ (((−1)^k )/k) =−(Σ_(k=1) ^∞ (((−1)^k )/k) +1) =−(−ln(2)+1) =ln(2)−1 Σ_(k=3) ^∞ (((−1)^(k−2) )/k) =Σ_(k=1) ^∞ (((−1)^k )/k) −(−1+(1/2))=−ln(2)+(1/2) Σ_(k=3) ^∞ (((−1)^(k−2) )/k^2 ) =Σ_(k=1) ^∞ (((−1)^k )/k^2 )−(−1+(1/2)) =−(π^2 /(12))+(1/2) ⇒ S=(1/4)(−(π^2 /(12)))−ln(2)+1−ln(2)+(1/2) +(3/4)(−(π^2 /(12))+(1/2)) =−(π^2 /(48))−2ln(2)+(3/2)+(3/8)−((3π^2 )/(48)) =((−4π^2 )/(48))−2ln(2)+((15)/8) =−(π^2 /(12))−2ln(2)+((15)/8)

$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)^{\mathrm{2}} }\left(−\mathrm{1}\right)^{{k}} \:\:\:{we}\:{have}\:{S}={lim}_{{n}\rightarrow+\infty} {S}_{{n}} \\ $$$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{{x}+\mathrm{2}}\:+\frac{{e}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${e}\:={lim}_{{x}\rightarrow−\mathrm{2}} \:\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} {F}\left({x}\right)\:=\frac{−\mathrm{3}}{−\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{{x}+\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+{c}+{d} \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{18}}\:=\frac{\mathrm{1}}{\mathrm{6}}\:={a}+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{{c}}{\mathrm{2}}\:+\frac{{d}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{12}}\:={a}+\frac{{c}}{\mathrm{2}}\:+\frac{{d}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{1}\:=\mathrm{6}{a}+\mathrm{3}{c}+\mathrm{2}{d}+\mathrm{2}\:\Rightarrow\mathrm{6}{a}+\mathrm{3}{c}+\mathrm{2}{d}=−\mathrm{1} \\ $$$${F}\left(−\mathrm{3}\right)\:=\frac{−\mathrm{5}}{\mathrm{9}\left(−\mathrm{2}\right)}\:=\frac{\mathrm{5}}{\mathrm{18}}\:=−\frac{{a}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{36}}−\frac{{c}}{\mathrm{2}}−{d}+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$=−\frac{{a}}{\mathrm{3}}−\frac{{c}}{\mathrm{2}}−{d}\:+\frac{\mathrm{7}}{\mathrm{9}}\:\Rightarrow\mathrm{5}=−\mathrm{6}{a}−\mathrm{9}{c}−\mathrm{18}{d}+\mathrm{14}\:\:\Rightarrow \\ $$$$−\mathrm{6}{a}−\mathrm{9}{c}−\mathrm{18}{d}\:=−\mathrm{9}\:\Rightarrow\mathrm{6}{a}+\mathrm{9}{c}+\mathrm{18}{d}\:=\mathrm{9}\:\Rightarrow\mathrm{2}{a}+\mathrm{3}{c}+\mathrm{6}{d}\:=\mathrm{3} \\ $$$${we}\:{get}\:{the}\:{system}\:\:\begin{cases}{{a}+{c}+{d}=\mathrm{0}}\\{\mathrm{6}{a}+\mathrm{3}{c}+\mathrm{2}{d}\:=−\mathrm{1}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\mathrm{2}{a}+\mathrm{3}{c}+\mathrm{6}{d}\:=\mathrm{3}\:\Rightarrow{d}=−{a}−{c}\:{and}\right. \\ $$$$\begin{cases}{\mathrm{6}{a}+\mathrm{3}{c}+\mathrm{2}\left(−{a}−{c}\right)=−\mathrm{1}}\\{\mathrm{2}{a}+\mathrm{3}{c}+\mathrm{6}\left(−{a}−{c}\right)=\mathrm{3}\:\Rightarrow\begin{cases}{\mathrm{4}{a}+{c}=−\mathrm{1}}\\{−\mathrm{4}{a}−\mathrm{3}{c}\:=\mathrm{3}\:\Rightarrow−\mathrm{2}{c}=\mathrm{2}\:\Rightarrow{c}=−\mathrm{1}}\end{cases}}\end{cases} \\ $$$$\mathrm{4}{a}=−\mathrm{1}+\mathrm{1}\:=\mathrm{0}\:\Rightarrow{a}=\mathrm{0}\:\Rightarrow{d}=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{{x}+\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{F}\left({k}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }−\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{2}} \\ $$$$+\frac{\mathrm{3}}{\mathrm{4}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{2}\right)^{\mathrm{2}} }\:{and}\:{lim}_{{n}\rightarrow+\infty} {S}_{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }−\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:+\sum_{{k}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{2}} }{{k}}+\frac{\mathrm{3}}{\mathrm{4}}\sum_{{k}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{2}} }{{k}^{\mathrm{2}} } \\ $$$$\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:=−\sum_{{k}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:=−\left(\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:+\mathrm{1}\right) \\ $$$$=−\left(−{ln}\left(\mathrm{2}\right)+\mathrm{1}\right)\:={ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{2}} }{{k}}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\:−\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=−{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{3}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{2}} }{{k}^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}^{\mathrm{2}} }−\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)−{ln}\left(\mathrm{2}\right)+\mathrm{1}−{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{4}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{48}}−\mathrm{2}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$=\frac{−\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{48}}−\mathrm{2}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{15}}{\mathrm{8}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{2}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{15}}{\mathrm{8}} \\ $$

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