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Question Number 60499 by abdo mathsup 649 cc last updated on 21/May/19
find the value of Σ_(n=1) ^∞ (((−1)^n )/(n^3 (n+1)^4 ))
$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$
Commented by maxmathsup by imad last updated on 30/May/19
let decompose F(x) =(1/(x^3 (x+1)^4 )) we have F(x) =Σ_(i=1) ^3 (a_i /x^i ) +Σ_(i=1) ^4 (b_i /((x+1)^i )) let determine D_2 (0) for f(x) =(x+1)^(−4) we have f(x)=f(0) +((f^′ (0))/(1!)) x +((f^((2)) (0))/(2!)) x^2 +(x^3 /(3!))ξ(x) f(0) =1 ,f^′ (x) =−4(x+1)^(−5) ⇒f^′ (0) =−4 ,f^((2)) (x)=20 (x+1)^(−6) ⇒ f^((2)) (0) =20 ⇒f(x)=1−4x +10x^2 +x^3 ξ(x) ⇒ ((f(x))/x^3 ) =(1/x^3 ) −(4/x^2 ) +((10)/x) +ξ(x) ⇒a_1 =10 , a_2 =−4 , a_3 =−1 also we have F(x) =_(x+1=t) (1/((t−1)^3 t^4 )) let find D_3 (0) for g(t) =(t−1)^(−3) g(t) =g(0) +((g^′ (0))/(1!)) t +((g^((2)) (0))/(2!)) t^2 +((g^((3)) (0))/(3!))t^3 +(t^4 /(4!))ξ(t) g(0) =−1 , g^′ (t) =−3(t−1)^(−4) ⇒g^′ (0) =−3 g^((2)) (t) =12(t−1)^(−5) ⇒g^((2)) (0) =−12 , g^((3)) (t) =−60(t−1)^(−6) ⇒ g^((3)) (0) =−60 ⇒ g(t) =−1 −3t −6t^2 −10 t^3 +(t^4 /(4!))ξ(t) ⇒ ((g(t))/t^4 ) =−(1/t^4 ) −(3/t^3 ) −(6/t^2 ) −((10)/t) +(1/(4!))ξ(t) =−(1/((x+1)^4 )) −(3/((x+1)^3 )) −(6/((x+1)^2 )) −((10)/((x+1))) +(1/(4!))ξ(x+1) ⇒ b_1 =−10 , b_2 =−6 , b_3 =−3 , b_4 =−1 and F(x) =((10)/x) −(4/x^2 ) −(1/x^3 ) −((10)/((x+1))) −(6/((x+1)^2 )) −(3/((x+1)^3 )) −(1/((x+1)^4 )) ⇒ Σ_(n=1) ^∞ (((−1)^n )/(n^3 (n+1)^4 )) =Σ_(n=1) ^∞ (−1)^n F(n) =10 Σ_(n=1) ^∞ (((−1)^n )/n) −4 Σ_(n=1) ^∞ (((−1)^n )/n^2 ) −10 Σ_(n=1) ^∞ (((−1)^n )/(n+1)) −6Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) −3 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) −Σ_(n=1) ^∞ (((−1)^n )/n^4 )
$${let}\:{decompose}\:{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{4}} }\:\:{we}\:{have}\: \\ $$$${F}\left({x}\right)\:=\sum_{{i}=\mathrm{1}} ^{\mathrm{3}} \:\frac{{a}_{{i}} }{{x}^{{i}} }\:+\sum_{{i}=\mathrm{1}} ^{\mathrm{4}} \:\frac{{b}_{{i}} }{\left({x}+\mathrm{1}\right)^{{i}} }\:\:{let}\:{determine}\:{D}_{\mathrm{2}} \left(\mathrm{0}\right)\:{for} \\ $$$${f}\left({x}\right)\:=\left({x}+\mathrm{1}\right)^{−\mathrm{4}} \:\:{we}\:{have}\:{f}\left({x}\right)={f}\left(\mathrm{0}\right)\:+\frac{{f}^{'} \left(\mathrm{0}\right)}{\mathrm{1}!}\:{x}\:+\frac{{f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)}{\mathrm{2}!}\:{x}^{\mathrm{2}} \:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\xi\left({x}\right) \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{1}\:\:,{f}^{'} \left({x}\right)\:=−\mathrm{4}\left({x}+\mathrm{1}\right)^{−\mathrm{5}} \:\Rightarrow{f}^{'} \left(\mathrm{0}\right)\:=−\mathrm{4}\:,{f}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{20}\:\left({x}+\mathrm{1}\right)^{−\mathrm{6}} \Rightarrow \\ $$$${f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{20}\:\Rightarrow{f}\left({x}\right)=\mathrm{1}−\mathrm{4}{x}\:+\mathrm{10}{x}^{\mathrm{2}} \:+{x}^{\mathrm{3}} \:\xi\left({x}\right)\:\Rightarrow \\ $$$$\frac{{f}\left({x}\right)}{{x}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{10}}{{x}}\:+\xi\left({x}\right)\:\Rightarrow{a}_{\mathrm{1}} =\mathrm{10}\:,\:{a}_{\mathrm{2}} =−\mathrm{4}\:,\:{a}_{\mathrm{3}} =−\mathrm{1}\:{also}\:{we}\:{have} \\ $$$${F}\left({x}\right)\:=_{{x}+\mathrm{1}={t}} \:\:\:\frac{\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} {t}^{\mathrm{4}} }\:\:{let}\:{find}\:\:{D}_{\mathrm{3}} \left(\mathrm{0}\right)\:\:{for}\:{g}\left({t}\right)\:=\left({t}−\mathrm{1}\right)^{−\mathrm{3}} \\ $$$${g}\left({t}\right)\:={g}\left(\mathrm{0}\right)\:+\frac{{g}^{'} \left(\mathrm{0}\right)}{\mathrm{1}!}\:{t}\:+\frac{{g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)}{\mathrm{2}!}\:{t}^{\mathrm{2}} \:+\frac{{g}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)}{\mathrm{3}!}{t}^{\mathrm{3}} \:+\frac{{t}^{\mathrm{4}} }{\mathrm{4}!}\xi\left({t}\right) \\ $$$${g}\left(\mathrm{0}\right)\:=−\mathrm{1}\:,\:{g}^{'} \left({t}\right)\:=−\mathrm{3}\left({t}−\mathrm{1}\right)^{−\mathrm{4}} \:\Rightarrow{g}^{'} \left(\mathrm{0}\right)\:=−\mathrm{3} \\ $$$${g}^{\left(\mathrm{2}\right)} \left({t}\right)\:=\mathrm{12}\left({t}−\mathrm{1}\right)^{−\mathrm{5}} \Rightarrow{g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=−\mathrm{12}\:\:,\:{g}^{\left(\mathrm{3}\right)} \left({t}\right)\:=−\mathrm{60}\left({t}−\mathrm{1}\right)^{−\mathrm{6}} \:\Rightarrow \\ $$$${g}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)\:=−\mathrm{60}\:\Rightarrow \\ $$$${g}\left({t}\right)\:=−\mathrm{1}\:−\mathrm{3}{t}\:−\mathrm{6}{t}^{\mathrm{2}} \:−\mathrm{10}\:{t}^{\mathrm{3}} \:+\frac{{t}^{\mathrm{4}} }{\mathrm{4}!}\xi\left({t}\right)\:\Rightarrow \\ $$$$\frac{{g}\left({t}\right)}{{t}^{\mathrm{4}} }\:=−\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\:−\frac{\mathrm{3}}{{t}^{\mathrm{3}} }\:−\frac{\mathrm{6}}{{t}^{\mathrm{2}} }\:−\frac{\mathrm{10}}{{t}}\:+\frac{\mathrm{1}}{\mathrm{4}!}\xi\left({t}\right) \\ $$$$=−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }\:−\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{6}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{10}}{\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{4}!}\xi\left({x}+\mathrm{1}\right)\:\Rightarrow \\ $$$${b}_{\mathrm{1}} =−\mathrm{10}\:,\:{b}_{\mathrm{2}} =−\mathrm{6}\:,\:{b}_{\mathrm{3}} =−\mathrm{3}\:,\:{b}_{\mathrm{4}} =−\mathrm{1}\:{and}\: \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{10}}{{x}}\:−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:−\frac{\mathrm{10}}{\left({x}+\mathrm{1}\right)}\:−\frac{\mathrm{6}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{4}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {F}\left({n}\right) \\ $$$$=\mathrm{10}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:−\mathrm{4}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:−\mathrm{10}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:−\mathrm{6}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$−\mathrm{3}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{4}} } \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 30/May/19
Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4)ξ(2) =(3/4) (π^2 /6) =(π^2 /8) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(24)) −(π^2 /8) =((−2π^2 )/(24)) =−(π^2 /(12)) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n) =−Σ_(n=2) ^∞ (((−1)^n )/n) =−{Σ_(n=1) ^∞ (((−1)^n )/n) +1) =−(−ln(2))−1 =ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^2 ) =−{ Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +1) =−(−(π^2 /(12)))−1 =(π^2 /(12)) −1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^3 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^3 ) for that let calculate δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) interms of ξ(x) =Σ_(n=1) ^∞ (1/n^x ) (x>1) we have δ(x) =Σ_(n=1) ^∞ (1/(2^x n^x )) − Σ_(n=0) ^∞ (1/((2n+1)^x )) ξ(x) =Σ_(n=1) ^∞ (1/n^x ) =(1/2^x ) ξ(x) +Σ_(n=0) ^∞ (1/((2n+1)^x )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^x )) =(1−2^(−x) )ξ(x) ⇒ δ(x) =2^(−x) ξ(x)−(1−2^(−x) )ξ(x) =(2^(1−x) −1)ξ(x) Σ_(n=1) ^∞ (((−1)^n )/n^3 ) =(2^(−2) −1)ξ(3)=((1/4)−1)ξ(3) =−(3/4)ξ(3) Σ_(n=1) ^∞ (((−1)^n )/n^4 ) =(2^(−3) −1)ξ(4) =((1/8) −1)ξ(4) =−(7/8)ξ(4) so the value of S is determined...
$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=\frac{−\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{24}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:=−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\boldsymbol{{n}}} \\ $$$$=−\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:+\mathrm{1}\right)\:=−\left(−{ln}\left(\mathrm{2}\right)\right)−\mathrm{1}\:={ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$=−\left\{\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:+\mathrm{1}\right)\:=−\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)−\mathrm{1}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} }\:=−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} } \\ $$$${for}\:{that}\:{let}\:\:{calculate}\:\delta\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:{interms}\:{of} \\ $$$$\xi\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{{x}} }\:\:\:\:\:\:\left({x}>\mathrm{1}\right)\:\:{we}\:{have}\: \\ $$$$\delta\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{2}^{{x}} {n}^{{x}} }\:−\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{x}} }\: \\ $$$$\xi\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{{x}} }\:=\frac{\mathrm{1}}{\mathrm{2}^{{x}} }\:\xi\left({x}\right)\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{x}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{x}} }\:=\left(\mathrm{1}−\mathrm{2}^{−{x}} \right)\xi\left({x}\right)\:\Rightarrow \\ $$$$\delta\left({x}\right)\:=\mathrm{2}^{−{x}} \xi\left({x}\right)−\left(\mathrm{1}−\mathrm{2}^{−{x}} \right)\xi\left({x}\right)\:=\left(\mathrm{2}^{\mathrm{1}−{x}} −\mathrm{1}\right)\xi\left({x}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} }\:=\left(\mathrm{2}^{−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{3}\right)=\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)\xi\left(\mathrm{3}\right)\:=−\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{4}} }\:=\left(\mathrm{2}^{−\mathrm{3}} −\mathrm{1}\right)\xi\left(\mathrm{4}\right)\:=\left(\frac{\mathrm{1}}{\mathrm{8}}\:−\mathrm{1}\right)\xi\left(\mathrm{4}\right)\:=−\frac{\mathrm{7}}{\mathrm{8}}\xi\left(\mathrm{4}\right) \\ $$$${so}\:{the}\:{value}\:{of}\:{S}\:{is}\:{determined}… \\ $$

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