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Question Number 82289 by mathmax by abdo last updated on 19/Feb/20
calculate Σ_(n=6) ^∞ (1/(n^2 −25))
$${calculate}\:\sum_{{n}=\mathrm{6}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{25}} \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
let S_n =Σ_(k=6) ^n (1/(k^2 −25)) ⇒S_n =(1/(10))Σ_(k=6) ^n {(1/(k−5))−(1/(k+5))} ⇒ 10 S_n =Σ_(k=6) ^n (1/(k−5))−Σ_(k=6) ^n (1/(k+5)) we have Σ_(k=6) ^n (1/(k−5)) =_(k−5=p) Σ_(p=1) ^(n−5) (1/p) =H_(n−5) Σ_(k=6) ^n (1/(k+5)) =_(k+5=p) Σ_(p=11) ^(n+5) (1/p) =Σ_(p=1) ^(n+5) (1/p)−(1+(1/2)+(1/3)+(1/4)+(1/5)+(1/6) +(1/7)+(1/8) +(1/9) +(1/(10)))=H_(n+5) −{(3/2)+(7/(12)) +((11)/(30)) +((15)/(56)) +((19)/(99))} ⇒ 10S_n =H_(n−5) −H_(n+5) +(3/2) +(7/(12)) +((11)/(30)) +((15)/(56)) +((19)/(99)) ∼ln(n−5)+γ +o((1/(n−5)))−γ−ln(n+5)−γ−o((1/(n+5))) +(3/2) +(7/(12)) +((11)/(30)) +((15)/(56)) +((19)/(99)) ⇒10S_n =ln(((n−5)/(n+5)))+(3/2)+(7/(12))+((11)/(30)) +((15)/(56))+((19)/(99)) ⇒lim_(n→+∞) S_n =(1/(10))((3/2) +(7/(12)) +((11)/(30))+((15)/(56))+((19)/(99)))
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{6}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{25}}\:\Rightarrow{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{10}}\sum_{{k}=\mathrm{6}} ^{{n}} \left\{\frac{\mathrm{1}}{{k}−\mathrm{5}}−\frac{\mathrm{1}}{{k}+\mathrm{5}}\right\}\:\Rightarrow \\ $$$$\mathrm{10}\:{S}_{{n}} =\sum_{{k}=\mathrm{6}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{5}}−\sum_{{k}=\mathrm{6}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{5}}\:{we}\:{have}\: \\ $$$$\sum_{{k}=\mathrm{6}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{5}}\:=_{{k}−\mathrm{5}={p}} \:\sum_{{p}=\mathrm{1}} ^{{n}−\mathrm{5}} \:\frac{\mathrm{1}}{{p}}\:={H}_{{n}−\mathrm{5}} \\ $$$$\sum_{{k}=\mathrm{6}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{5}}\:=_{{k}+\mathrm{5}={p}} \:\:\sum_{{p}=\mathrm{11}} ^{{n}+\mathrm{5}} \:\frac{\mathrm{1}}{{p}}\:=\sum_{{p}=\mathrm{1}} ^{{n}+\mathrm{5}} \frac{\mathrm{1}}{{p}}−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}\right. \\ $$$$\left.+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{9}}\:+\frac{\mathrm{1}}{\mathrm{10}}\right)={H}_{{n}+\mathrm{5}} −\left\{\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{7}}{\mathrm{12}}\:+\frac{\mathrm{11}}{\mathrm{30}}\:+\frac{\mathrm{15}}{\mathrm{56}}\:+\frac{\mathrm{19}}{\mathrm{99}}\right\}\:\Rightarrow \\ $$$$\mathrm{10}{S}_{{n}} ={H}_{{n}−\mathrm{5}} −{H}_{{n}+\mathrm{5}} +\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{7}}{\mathrm{12}}\:+\frac{\mathrm{11}}{\mathrm{30}}\:+\frac{\mathrm{15}}{\mathrm{56}}\:+\frac{\mathrm{19}}{\mathrm{99}} \\ $$$$\sim{ln}\left({n}−\mathrm{5}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}−\mathrm{5}}\right)−\gamma−{ln}\left({n}+\mathrm{5}\right)−\gamma−{o}\left(\frac{\mathrm{1}}{{n}+\mathrm{5}}\right) \\ $$$$+\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{7}}{\mathrm{12}}\:+\frac{\mathrm{11}}{\mathrm{30}}\:+\frac{\mathrm{15}}{\mathrm{56}}\:+\frac{\mathrm{19}}{\mathrm{99}}\:\Rightarrow\mathrm{10}{S}_{{n}} ={ln}\left(\frac{{n}−\mathrm{5}}{{n}+\mathrm{5}}\right)+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{7}}{\mathrm{12}}+\frac{\mathrm{11}}{\mathrm{30}}\:+\frac{\mathrm{15}}{\mathrm{56}}+\frac{\mathrm{19}}{\mathrm{99}} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{10}}\left(\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{7}}{\mathrm{12}}\:+\frac{\mathrm{11}}{\mathrm{30}}+\frac{\mathrm{15}}{\mathrm{56}}+\frac{\mathrm{19}}{\mathrm{99}}\right) \\ $$$$ \\ $$
Commented by mr W last updated on 20/Feb/20
typo in ((19)/(99)). should be ((19)/(90)).
$${typo}\:{in}\:\:\frac{\mathrm{19}}{\mathrm{99}}.\:{should}\:{be}\:\frac{\mathrm{19}}{\mathrm{90}}. \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
yes sir you are right thanks
$${yes}\:{sir}\:{you}\:{are}\:{right}\:{thanks} \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
forgive error of typo change 99 by 90 ⇒ lim_(n→+∞) S_n =(1/(10))((3/2)+(7/(12)) +((11)/(30)) +((15)/(56)) +((19)/(90)))
$${forgive}\:{error}\:{of}\:{typo}\:{change}\:\mathrm{99}\:{by}\:\mathrm{90}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{10}}\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{7}}{\mathrm{12}}\:+\frac{\mathrm{11}}{\mathrm{30}}\:+\frac{\mathrm{15}}{\mathrm{56}}\:+\frac{\mathrm{19}}{\mathrm{90}}\right) \\ $$
Answered by mr W last updated on 20/Feb/20
Σ_(n=6) ^∞ (1/(n^2 −25)) =(1/(10))Σ_(n=6) ^∞ ((1/(n−5))−(1/(n+5))) =(1/(10)){Σ_(n=6) ^∞ (1/(n−5))−Σ_(n=6) ^∞ (1/(n+5))} =(1/(10)){Σ_(n=1) ^∞ (1/n)−Σ_(n=10) ^∞ (1/(n+1))} =(1/(10)){Σ_(n=1) ^∞ (1/n)−Σ_(n=1) ^∞ (1/(n+1))+Σ_(n=1) ^9 (1/(n+1))} =(1/(10)){Σ_(n=1) ^∞ ((1/n)−(1/(n+1)))}+(1/(10))((1/2)+(1/3)+...+(1/(10))) =(1/(10))(1)+(1/(10))((1/2)+(1/3)+...+(1/(10))) =(1/(10))(1+(1/2)+(1/3)+...+(1/(10))) =((7381)/(25200))≈0.2929
$$\underset{{n}=\mathrm{6}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{25}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\underset{{n}=\mathrm{6}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\mathrm{5}}−\frac{\mathrm{1}}{{n}+\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left\{\underset{{n}=\mathrm{6}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\mathrm{5}}−\underset{{n}=\mathrm{6}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{5}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{n}=\mathrm{10}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}+\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\right\}+\frac{\mathrm{1}}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{10}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{10}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{10}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{10}}\right) \\ $$$$=\frac{\mathrm{7381}}{\mathrm{25200}}\approx\mathrm{0}.\mathrm{2929} \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
thank you sir mrw.
$${thank}\:{you}\:{sir}\:{mrw}. \\ $$

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