Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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)))

$${let}\:{S}\:=\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{{n}}{\left({n}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} }\:\Rightarrow\:{S}\:=\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{{n}}{\left({n}−\mathrm{3}\right)^{\mathrm{2}} \left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${but}\:\:\left({n}+\mathrm{3}\right)^{\mathrm{2}} −\left({n}−\mathrm{3}\right)^{\mathrm{2}} \:={n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{9}−{n}^{\mathrm{2}} +\mathrm{6}{n}−\mathrm{9}\:=\mathrm{12}{n}\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{\left({n}+\mathrm{3}\right)^{\mathrm{2}} −\left({n}−\mathrm{3}\right)^{\mathrm{2}} }{\left({n}+\mathrm{3}\right)^{\mathrm{2}} \left({n}−\mathrm{3}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}−\mathrm{3}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{4}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\:\:\:\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{7}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{12}}\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}\:+\frac{\mathrm{1}}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{25}}+\frac{\mathrm{1}}{\mathrm{36}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com