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Question Number 143254 by Mathspace last updated on 12/Jun/21
$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$
Answered by Olaf_Thorendsen last updated on 12/Jun/21
![R(n) = (1/(n^2 (n+1)(n+2)(n+3))) R(n) = (1/(6n^2 ))−((11)/(36n))+(1/(2(n+1)))−(1/(4(n+2)))+(1/(18(n+3))) Let = A_n = Σ_(n=1) ^∞ (((−1)^n )/(n+1)) = ln2 Let = B_n = Σ_(n=1) ^∞ (((−1)^n )/n^2 ) = −(π^2 /(12)) Let = S_n = Σ_(n=1) ^∞ (−1)^n R(n) S_n = Σ_(n=1) ^∞ (((−1)^n )/(6n^2 ))−Σ_(n=1) ^∞ ((11(−1)^n )/(36n))+Σ_(n=1) ^∞ (((−1)^n )/(2(n+1))) −Σ_(n=1) ^∞ (((−1)^n )/(4(n+2)))+Σ_(n=1) ^∞ (((−1)^n )/(18(n+3))) S_n = (1/6)B_n −((11)/(36))[−1−A_n ]+(1/2)A_n −(1/4)[−(1/2)−A_n ]+(1/(18))[(1/2)−(1/3)+A_n ] S_n = ((11)/(36))+(1/8)+(1/(36))−(1/(54))+(1/6)B_n +(((11)/(36))+(1/2)+(1/4)+(1/(18)))A_n S_n = ((95)/(216))+((10)/9)A_n +(1/6)B_n S_n = ((95)/(216))+((10)/9)ln2+(1/6)(−(π^2 /(12))) S_n = ((95)/(216))+((10)/9)ln2−(π^2 /(72))](Q143307.png)
$${R}\left({n}\right)\:=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$${R}\left({n}\right)\:=\:\frac{\mathrm{1}}{\mathrm{6}{n}^{\mathrm{2}} }−\frac{\mathrm{11}}{\mathrm{36}{n}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{18}\left({n}+\mathrm{3}\right)} \\ $$$$\mathrm{Let}\:=\:{A}_{{n}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:=\:\mathrm{ln2} \\ $$$$\mathrm{Let}\:=\:{B}_{{n}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\mathrm{Let}\:=\:{S}_{{n}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {R}\left({n}\right) \\ $$$${S}_{{n}} \:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{6}{n}^{\mathrm{2}} }−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{11}\left(−\mathrm{1}\right)^{{n}} }{\mathrm{36}{n}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}\left({n}+\mathrm{2}\right)}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{18}\left({n}+\mathrm{3}\right)} \\ $$$${S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{6}}{B}_{{n}} −\frac{\mathrm{11}}{\mathrm{36}}\left[−\mathrm{1}−{A}_{{n}} \right]+\frac{\mathrm{1}}{\mathrm{2}}{A}_{{n}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\left[−\frac{\mathrm{1}}{\mathrm{2}}−{A}_{{n}} \right]+\frac{\mathrm{1}}{\mathrm{18}}\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+{A}_{{n}} \right] \\ $$$${S}_{{n}} \:=\:\frac{\mathrm{11}}{\mathrm{36}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{36}}−\frac{\mathrm{1}}{\mathrm{54}}+\frac{\mathrm{1}}{\mathrm{6}}{B}_{{n}} \\ $$$$+\left(\frac{\mathrm{11}}{\mathrm{36}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{18}}\right){A}_{{n}} \\ $$$${S}_{{n}} \:=\:\frac{\mathrm{95}}{\mathrm{216}}+\frac{\mathrm{10}}{\mathrm{9}}{A}_{{n}} +\frac{\mathrm{1}}{\mathrm{6}}{B}_{{n}} \\ $$$${S}_{{n}} \:=\:\frac{\mathrm{95}}{\mathrm{216}}+\frac{\mathrm{10}}{\mathrm{9}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{6}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right) \\ $$$${S}_{{n}} \:=\:\frac{\mathrm{95}}{\mathrm{216}}+\frac{\mathrm{10}}{\mathrm{9}}\mathrm{ln2}−\frac{\pi^{\mathrm{2}} }{\mathrm{72}} \\ $$