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Question Number 116555 by Bird last updated on 04/Oct/20
find Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)^3 (n+2)^4 ))
$${find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{2}\right)^{\mathrm{4}} } \\ $$
Answered by Olaf last updated on 05/Oct/20
R(n) = (1/(n^2 (n+1)^3 (n+2)^4 )) R(n) = −(5/(16)).(1/n)+(1/(16)).(1/n^2 )+(5/(n+1))−(2/((n+1)^2 )) +(1/((n+1)^3 ))−((75)/(16)).(1/(n+2))−((39)/(16)).(1/((n+2)^2 )) −(1/((n+2)^3 ))−(1/4).(1/((n+2)^4 )) Σ_(n=1) ^∞ (((−1)^n )/n) = −ln2 Σ_(n=1) ^∞ (((−1)^n )/n^2 ) = −(π^2 /(12)) Σ_(n=1) ^∞ (((−1)^n )/n^3 ) = −(3/(12))ξ(3) Σ_(n=1) ^∞ (((−1)^n )/n^4 ) = −((7π^4 )/(720)) S = −(5/(16))(−ln2)+(1/(16))(−(π^2 /(12)))+5(−ln2+1) −2(−(π^2 /(12))+1)+(−(3/(12))ξ(3)+1)−((75)/(16))(−ln2+1−(1/2)) −((39)/(16))(−(π^2 /(12))+1−(1/4))−(−(3/(12))ξ(3)+1−(1/8)) −(1/4)(−((7π^4 )/(720))+1−(1/(16))) S = ln2((5/(16))−5+((75)/(16))) −(π^2 /(12))((1/(16))−2−((39)/(16))) −(3/(12))ξ(3)(1−1)+((7π^4 )/(2880)) +5−2+1−((75)/(32))−((117)/(64))−(7/8)−((15)/(64)) S = ((35π^2 )/(96))+((7π^4 )/(2880))−((41)/(32)) Please mister verify the calculous.
$$\mathrm{R}\left({n}\right)\:=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{2}\right)^{\mathrm{4}} } \\ $$$$\mathrm{R}\left({n}\right)\:=\:−\frac{\mathrm{5}}{\mathrm{16}}.\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{16}}.\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{5}}{{n}+\mathrm{1}}−\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{75}}{\mathrm{16}}.\frac{\mathrm{1}}{{n}+\mathrm{2}}−\frac{\mathrm{39}}{\mathrm{16}}.\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{4}} } \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=\:−\mathrm{ln2} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} }\:=\:−\frac{\mathrm{3}}{\mathrm{12}}\xi\left(\mathrm{3}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{4}} }\:=\:−\frac{\mathrm{7}\pi^{\mathrm{4}} }{\mathrm{720}} \\ $$$$ \\ $$$$\mathrm{S}\:=\:−\frac{\mathrm{5}}{\mathrm{16}}\left(−\mathrm{ln2}\right)+\frac{\mathrm{1}}{\mathrm{16}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)+\mathrm{5}\left(−\mathrm{ln2}+\mathrm{1}\right) \\ $$$$−\mathrm{2}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\mathrm{1}\right)+\left(−\frac{\mathrm{3}}{\mathrm{12}}\xi\left(\mathrm{3}\right)+\mathrm{1}\right)−\frac{\mathrm{75}}{\mathrm{16}}\left(−\mathrm{ln2}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$−\frac{\mathrm{39}}{\mathrm{16}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\left(−\frac{\mathrm{3}}{\mathrm{12}}\xi\left(\mathrm{3}\right)+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\mathrm{7}\pi^{\mathrm{4}} }{\mathrm{720}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right) \\ $$$$\mathrm{S}\:=\:\mathrm{ln2}\left(\frac{\mathrm{5}}{\mathrm{16}}−\mathrm{5}+\frac{\mathrm{75}}{\mathrm{16}}\right) \\ $$$$−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\left(\frac{\mathrm{1}}{\mathrm{16}}−\mathrm{2}−\frac{\mathrm{39}}{\mathrm{16}}\right) \\ $$$$−\frac{\mathrm{3}}{\mathrm{12}}\xi\left(\mathrm{3}\right)\left(\mathrm{1}−\mathrm{1}\right)+\frac{\mathrm{7}\pi^{\mathrm{4}} }{\mathrm{2880}} \\ $$$$+\mathrm{5}−\mathrm{2}+\mathrm{1}−\frac{\mathrm{75}}{\mathrm{32}}−\frac{\mathrm{117}}{\mathrm{64}}−\frac{\mathrm{7}}{\mathrm{8}}−\frac{\mathrm{15}}{\mathrm{64}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{35}\pi^{\mathrm{2}} }{\mathrm{96}}+\frac{\mathrm{7}\pi^{\mathrm{4}} }{\mathrm{2880}}−\frac{\mathrm{41}}{\mathrm{32}} \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{mister}\:\mathrm{verify}\:\mathrm{the}\:\mathrm{calculous}. \\ $$
Commented by mathmax by abdo last updated on 05/Oct/20
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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