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Question Number 79374 by TawaTawa last updated on 24/Jan/20

Answered by mind is power last updated on 24/Jan/20

cv ∼(1/n^2 ) Riemann ((n−1)/((n+1)(n+2)(n+3)))=(1/(n+1))+((−3)/(−1.1.(n+2)))−(4/((−2)(−1)(n+3))) =Σ_(n≥1) {−(1/(n+1))+(3/((n+2)))−(2/((n+3)))} =Σ_(n≥1) (−(1/(n+1))+(1/(n+2)))+2Σ((1/(n+2))−(1/(n+3))) Σ_(n≥1) (−(1/(n+1))+(1/(n+2)))=Σ_(n≥1) (−(1/2)+(1/3)−(1/3)+(1/4)−(1/4)+.......)=−(1/2) Σ((1/(n+2))−(1/(n+3)))=((1/3)−(1/4)+(1/4)−(1/5)+.....)=(1/3) =−(1/2)+2.(1/3)=(1/6)=Σ_(n≥1) ((n−1)/((n+1)(n+2)(n+3)))

$${cv}\:\sim\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:{Riemann} \\ $$$$\frac{{n}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{−\mathrm{3}}{−\mathrm{1}.\mathrm{1}.\left({n}+\mathrm{2}\right)}−\frac{\mathrm{4}}{\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\left\{−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{3}}{\left({n}+\mathrm{2}\right)}−\frac{\mathrm{2}}{\left({n}+\mathrm{3}\right)}\right\} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)+\mathrm{2}\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}+.......\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Sigma\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)=\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}+.....\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}.\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{n}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$

Commented by TawaTawa last updated on 24/Jan/20

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by mind is power last updated on 24/Jan/20

y′re Welcom you study What sir ?

$${y}'{re}\:{Welcom}\:{you}\:{study}\:{What}\:{sir}\:? \\ $$

Commented by TawaTawa last updated on 25/Jan/20

Am a girl

$$\mathrm{Am}\:\mathrm{a}\:\mathrm{girl} \\ $$

Commented by TawaTawa last updated on 25/Jan/20

Studying computer engineering sir

$$\mathrm{Studying}\:\:\mathrm{computer}\:\mathrm{engineering}\:\mathrm{sir} \\ $$

Commented by mind is power last updated on 25/Jan/20

so sorry ,my english is bad, Good luck for y′re Study ∴all you need is practice∴

$${so}\:{sorry}\:,{my}\:{english}\:{is}\:{bad}, \\ $$$${Good}\:{luck}\:\:{for}\:{y}'{re}\:{Study}\:\therefore{all}\:{you}\:{need}\:{is}\:{practice}\therefore \\ $$

Commented by TawaTawa last updated on 25/Jan/20

Thank you sir. God bless you.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Commented by TawaTawa last updated on 25/Jan/20

Sir, please help me solve Q79360

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\:\mathrm{Q79360} \\ $$

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