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Question Number 67010 by mathmax by abdo last updated on 21/Aug/19
calculate Σ_(n=4) ^(+∞) (n/((n^2 −9)^2 ))
$${calculate}\:\:\sum_{{n}=\mathrm{4}} ^{+\infty} \:\:\:\:\frac{{n}}{\left({n}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 22/Aug/19
let S =Σ_(n=4) ^∞ (n/((n^2 −9)^2 )) ⇒ S =Σ_(n=4) ^∞ (n/((n−3)^2 (n+3)^2 )) but (n+3)^2 −(n−3)^2 =n^2 +6n+9−n^2 +6n−9 =12n ⇒ S =(1/(12))Σ_(n=4) ^∞ (((n+3)^2 −(n−3)^2 )/((n+3)^2 (n−3)^2 )) =(1/(12))Σ_(n=4) ^∞ (1/((n−3)^2 ))−(1/(12))Σ_(n=4) ^∞ (1/((n+3)^2 )) = (1/(12))Σ_(n=1) ^∞ (1/n^2 )−(1/(12))Σ_(n=7) ^∞ (1/n^2 ) =(1/(12))Σ_(n=1) ^∞ (1/n^2 )−(1/(12)){Σ_(n=1) ^∞ (1/n^2 )−1−(1/2^2 )−(1/3^2 )−(1/4^2 )−(1/5^2 )−(1/6^2 )} =(1/(12))(1+(1/4)+(1/9) +(1/(16)) +(1/(25))+(1/(36)))
$${let}\:{S}\:=\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{{n}}{\left({n}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} }\:\Rightarrow\:{S}\:=\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{{n}}{\left({n}−\mathrm{3}\right)^{\mathrm{2}} \left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${but}\:\:\left({n}+\mathrm{3}\right)^{\mathrm{2}} −\left({n}−\mathrm{3}\right)^{\mathrm{2}} \:={n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{9}−{n}^{\mathrm{2}} +\mathrm{6}{n}−\mathrm{9}\:=\mathrm{12}{n}\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{\left({n}+\mathrm{3}\right)^{\mathrm{2}} −\left({n}−\mathrm{3}\right)^{\mathrm{2}} }{\left({n}+\mathrm{3}\right)^{\mathrm{2}} \left({n}−\mathrm{3}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}−\mathrm{3}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\:\:\:\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{7}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{12}}\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}\:+\frac{\mathrm{1}}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{25}}+\frac{\mathrm{1}}{\mathrm{36}}\right) \\ $$

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