Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 26121 by abdo imad last updated on 20/Dec/17

answer to26109 S_n = Σ_(k=1) ^(k=n) (1/(k^2 (k+1)^2 )) we decompose F(X) = (1/(X^2 (X+1)^2 )) = (a/X) +(b/X^2^ ) +(c/(X+1)) +(d/((X+1)^2 )) we find F(X) = ((−2)/X) +(1/X^2 ) + (2/(X+1)) + (1/((X+1)^2 )) so S_n = −2Σ_(k=1) ^(k=n) (1/k) +2Σ_(k=1) ^(k=n) (1/(k+1)) +Σ_(k=1) ^(k=n) (1/k^2^ ) + Σ_(k=1) ^(k=n) (1/((k+1)^2 )) but Σ_(k=1) ^(k=n) (1/k) = H_n Σ_(k=1) ^(k=n) (1/(k+1))= H_(n+1) −1 Σ_(k=1) ^(k=n) (1/((k+1)^2 )) = Σ_(k=1) ^(k=n) (1/k^2 ) + (1/((n+1)^2 )) −1⇒ S_n = 2(H_(n+1) −H_n ) +2 Σ_(k=1) ^(k=n) (1/k^2 ) −3 but lim_(n−>∝) (H_(n+1) − H_n ) =0 and Σ_(k=1) ^∝ (1/k^2 ) = (π^2 /6) ⇒ lim_(n−>∝) S_n = 2 (π^2 /6) −3 = (π^2 /3) −3 .

$${answer}\:{to}\mathrm{26109}\:\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{decompose} \\ $$$${F}\left({X}\right)\:\:=\:\:\frac{\mathrm{1}}{{X}^{\mathrm{2}} \left({X}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\:\frac{{a}}{{X}}\:\:\:+\frac{{b}}{{X}^{\mathrm{2}^{} } }\:\:+\frac{{c}}{{X}+\mathrm{1}}\:\:+\frac{{d}}{\left({X}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{find} \\ $$$${F}\left({X}\right)\:\:=\:\:\frac{−\mathrm{2}}{{X}}\:\:+\frac{\mathrm{1}}{{X}^{\mathrm{2}} }\:\:+\:\frac{\mathrm{2}}{{X}+\mathrm{1}}\:\:+\:\:\frac{\mathrm{1}}{\left({X}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{so} \\ $$$$\:\:{S}_{{n}} \:\:=\:\:−\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}}\:\:+\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:\:+\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}^{} } }\:\:+\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${but}\:\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}}\:\:=\:{H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}=\:\:{H}_{{n}+\mathrm{1}} \:\:−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:\:=\:\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:+\:\:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:−\mathrm{1}\Rightarrow \\ $$$${S}_{{n}} \:\:=\:\:\mathrm{2}\left({H}_{{n}+\mathrm{1}} \:\:−{H}_{{n}} \right)\:\:+\mathrm{2}\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:\:−\mathrm{3} \\ $$$${but}\:\:\:{lim}_{{n}−>\propto} \left({H}_{{n}+\mathrm{1}} −\:{H}_{{n}} \right)\:\:=\mathrm{0}\:\:\:{and}\:\:\:\:\sum_{{k}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\:\:\:{lim}_{{n}−>\propto} \:\:{S}_{{n}} \:\:=\:\mathrm{2}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:−\mathrm{3}\:\:=\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:\:−\mathrm{3}\:\:\:\:\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com