Question Number 41236 by maxmathsup by imad last updated on 04/Aug/18
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{1}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Aug/18
![(n+1)^3 −(n−1)^3 =(n^3 +3n^2 +3n+1)−(n^3 −3n^2 +3n−1) =6^ n^2 +2 Σ_(n=2) ^∞ (1/2).(((n+1)^3 −(n−1)^3 )/((n+1)^3 (n−1)^3 )) =(1/2)Σ_(n=2) ^∞ (1/((n−1)^3 ))−(1/((n+1)^3 ))=(1/2)Σ_2 ^∞ T_n T_n =[(1/((n−1)^3 )) −(1/((n+1)^3 ))] T_2 =(1/1^3 )−(1/3^3 ) T_3 =(1/2^3 )−(1/4^3 ) T_4 =(1/3^3 )−(1/5^3 ) T_5 =(1/4^3 )−(1/6^3 ) .... .... T_n =(1/((n−1)^3 ))−(1/((n+1)^3 )) s=(1/1^3 )+(1/2^3 )−(1/((n+1)^3 )) (1/2)Σ_2 ^∞ T_n =(1/2)(1+(1/8))−lim_(n→∞) (1/((n+1)^3 ))=(9/(16))−0=(9/(16)) ]](https://www.tinkutara.com/question/Q41260.png)
$$\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\left({n}−\mathrm{1}\right)^{\mathrm{3}} =\left({n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right)−\left({n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{1}\right) \\ $$$$=\mathrm{6}^{} {n}^{\mathrm{2}} +\mathrm{2} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}.\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\left({n}−\mathrm{1}\right)^{\mathrm{3}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} \left({n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{2}} {\overset{\infty} {\sum}}{T}_{{n}} \\ $$$${T}_{{n}} =\left[\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right] \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} } \\ $$$${T}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} } \\ $$$${T}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} } \\ $$$${T}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{3}} } \\ $$$$…. \\ $$$$…. \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{2}} {\overset{\infty} {\sum}}{T}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{8}}\right)−\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{9}}{\mathrm{16}}−\mathrm{0}=\frac{\mathrm{9}}{\mathrm{16}} \\ $$$$ \\ $$$$\left.\right] \\ $$
Commented by prof Abdo imad last updated on 04/Aug/18
$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{Tanmay}\:{thanks} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Aug/18
$${thank}\:{you}\:{sir}… \\ $$
Answered by prof Abdo imad last updated on 04/Aug/18
$${S}\:={lim}_{{n}\rightarrow+\infty} \:{S}_{{N}} \:\:{with}\:{S}_{{N}} =\sum_{{n}=\mathrm{2}} ^{{N}} \:\frac{\mathrm{3}{n}^{\mathrm{2}\:} +\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${but}\:{we}\:{have}\:\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\left({n}−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$=\mathrm{2}\left(\left({n}+\mathrm{1}\right)^{\mathrm{2}} \:+\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)\:+\left({n}−\mathrm{1}\right)^{\mathrm{3}} \right) \\ $$$$=\mathrm{2}\left({n}^{\mathrm{2}} \:+\mathrm{2}{n}+\mathrm{1}\:+{n}^{\mathrm{2}} −\mathrm{1}\:+{n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$=\mathrm{2}\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}\right)\Rightarrow{S}_{{N}} =\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{2}} ^{{N}} \frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\left({n}−\mathrm{1}\right)^{\mathrm{3}} }{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\sum_{{n}=\mathrm{2}} ^{{N}} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:−\sum_{{n}=\mathrm{2}} ^{{N}} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right\}\:\:{but} \\ $$$$\sum_{{n}=\mathrm{2}} ^{{N}} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{1}} ^{{N}−\mathrm{1}} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\xi_{{N}−\mathrm{1}} \left(\mathrm{3}\right) \\ $$$$\sum_{{n}=\mathrm{2}} ^{{N}} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{3}} ^{{N}+\mathrm{1}} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\xi_{{N}+\mathrm{1}} \left(\mathrm{3}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$${S}_{{N}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\xi_{{N}−\mathrm{1}} \left(\mathrm{3}\right)−\xi_{{N}+\mathrm{1}} \left(\mathrm{3}\right)\:+\frac{\mathrm{9}}{\mathrm{8}}\right\}{but} \\ $$$${lim}_{{N}\rightarrow+\infty} \xi_{{N}−\mathrm{1}} \left(\mathrm{3}\right)−\xi_{{N}+\mathrm{1}} \left(\mathrm{3}\right)=\xi\left(\mathrm{3}\right)−\xi\left(\mathrm{3}\right)=\mathrm{0}\Rightarrow \\ $$$${lim}_{{N}\rightarrow+\infty} \:{S}_{{N}} =\:\frac{\mathrm{9}}{\mathrm{16}}\:={S} \\ $$