Menu Close

find-the-value-of-n-1-2-n-3-3n-2-2n-

Tinku Tara 0 Comments
Pin




Question Number 33591 by abdo imad last updated on 19/Apr/18
find the value of Σ_(n=1) ^∞ (2/(n^3 +3n^2 +2n)) .
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{2}}{{n}^{\mathrm{3}} \:\:+\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{2}{n}}\:. \\ $$
Commented by abdo imad last updated on 20/Apr/18
let put S_n = Σ_(k=1) ^n (2/(k^3 +3k^2 +2k)) we have k^3 +3k^2 +2k =k( k^2 +3k +2)=k( k^2 +k +2k+2) =k (k(k+1) +2(k+1))=k(k+1)(k+2) let decompose F(x)= (2/(x(x+1)(x+2))) = (a/x) +(b/(x+1)) +(c/(x+2)) a =lim_(x→0) xF(x) = 1 b =lim_(x→−1) (x+1)F(x)= (2/((−1).1)) =−2 c =lim_(x→−2) (x+2)F(x) = (2/((−2)(−1))) =1 ⇒ F(x)=(1/x) −(2/(x+1)) +(1/(x+2)) and S_n = Σ_(k=1) ^n ( (1/k) −(2/(k+1)) +(1/(k+2)))=Σ_(k=1) ^n (1/k) −2Σ_(k=1) ^n (1/(k+1)) +Σ_(k=1) ^n (1/(k+2)) Σ_(k=1) ^n (1/k) = H_n Σ_(k=1) ^n (1/(k+1)) =Σ_(k=2) ^(n+1) (1/k) =H_(n+1) −1 Σ_(k=1) ^n (1/(k+2)) =Σ_(k=3) ^(n+2) (1/k) = H_(n+2) −(3/2) ⇒ S_n = H_n −2( H_(n+1) −1) +H_(n+2) −(3/2) S_n =H_n +H_(n+2) −2H_(n+1) +2 −(3/2) =H_n +H_(n+2) −2H_(n+1) +(1/2) but H_n = ln(n) +γ +o((1/n)) H_(n+2) =ln(n+2) +γ +o((1/n)) H_(n+1) = ln(n+1) +γ +o((1/n)) ⇒ S_n =ln(n(n+2)) +2γ −2ln(n+1) −2γ +o((1/n)) +(1/2) =ln(((n^2 +2n)/((n+1)^2 ))) +(1/2) ⇒ lim_(n→+∞) S_n =(1/2) ⇒ ★Σ_(n=1) ^∞ (2/(n^3 +3n^2 +2n)) = (1/2) ★
$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{2}}{{k}^{\mathrm{3}} \:+\mathrm{3}{k}^{\mathrm{2}} \:+\mathrm{2}{k}}\:\:{we}\:{have} \\ $$$${k}^{\mathrm{3}} \:+\mathrm{3}{k}^{\mathrm{2}} \:+\mathrm{2}{k}\:={k}\left(\:{k}^{\mathrm{2}} \:+\mathrm{3}{k}\:+\mathrm{2}\right)={k}\left(\:{k}^{\mathrm{2}} \:+{k}\:+\mathrm{2}{k}+\mathrm{2}\right) \\ $$$$={k}\:\left({k}\left({k}+\mathrm{1}\right)\:+\mathrm{2}\left({k}+\mathrm{1}\right)\right)={k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{2}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}\:=\:\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{{x}+\mathrm{2}} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} {xF}\left({x}\right)\:=\:\mathrm{1} \\ $$$${b}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\:\frac{\mathrm{2}}{\left(−\mathrm{1}\right).\mathrm{1}}\:=−\mathrm{2} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{2}} \left({x}+\mathrm{2}\right){F}\left({x}\right)\:=\:\frac{\mathrm{2}}{\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)}\:=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{2}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{{x}+\mathrm{2}}\:\:{and} \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(\:\frac{\mathrm{1}}{{k}}\:−\frac{\mathrm{2}}{{k}+\mathrm{1}}\:+\frac{\mathrm{1}}{{k}+\mathrm{2}}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}+\mathrm{1}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}+\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:=\:{H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}+\mathrm{1}} \:\:−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{1}}{{k}}\:\:\:=\:{H}_{{n}+\mathrm{2}} \:−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$${S}_{{n}} \:=\:{H}_{{n}} \:−\mathrm{2}\left(\:{H}_{{n}+\mathrm{1}} \:−\mathrm{1}\right)\:+{H}_{{n}+\mathrm{2}} \:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${S}_{{n}} \:={H}_{{n}} \:+{H}_{{n}+\mathrm{2}} \:−\mathrm{2}{H}_{{n}+\mathrm{1}} \:\:+\mathrm{2}\:−\frac{\mathrm{3}}{\mathrm{2}}\:={H}_{{n}} \:+{H}_{{n}+\mathrm{2}} \:−\mathrm{2}{H}_{{n}+\mathrm{1}} \:+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${but}\:{H}_{{n}} =\:{ln}\left({n}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$${H}_{{n}+\mathrm{2}} \:={ln}\left({n}+\mathrm{2}\right)\:+\gamma\:\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$${H}_{{n}+\mathrm{1}} \:=\:{ln}\left({n}+\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow \\ $$$${S}_{{n}} ={ln}\left({n}\left({n}+\mathrm{2}\right)\right)\:+\mathrm{2}\gamma\:\:−\mathrm{2}{ln}\left({n}+\mathrm{1}\right)\:−\mathrm{2}\gamma\:\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$={ln}\left(\frac{{n}^{\mathrm{2}} \:+\mathrm{2}{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} {S}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\bigstar\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{2}}{{n}^{\mathrm{3}} \:+\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{2}{n}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\bigstar \\ $$
Answered by sma3l2996 last updated on 19/Apr/18
Σ_(n=1) ^∞ (2/(n^3 +3n^2 +2n))=Σ_(n=1) ^∞ (2/(n(n+1)(n+2))) (1/(n(n+1)(n+2)))=(a/n)+(b/(n+1))+(c/(n+2)) a=(1/2) ; b=−1 ; c=(1/2) Σ_(n=1) ^∞ (2/(n(n+1)(n+2)))=Σ_(n=1) ^∞ ((1/n)−(2/(n+1))+(1/(n+2))) =Σ_(n=1) ^∞ (1/n)−Σ_(n=1) ^∞ (2/(n+1))+Σ_(n=1) ^∞ (1/(n+2)) let l=n+1 and k=n+2 Σ_(n=1) ^∞ (2/(n(n+1)(n+2)))=Σ_(n=1) ^∞ (1/n)−Σ_(l=2) ^∞ (2/l)+Σ_(k=3) ^∞ (1/k) =Σ_(n=1) ^∞ (1/n)−(Σ_(l=1) ^∞ (2/l)−(2/1))+(Σ_(k=1) ^∞ (1/k)−(1/1)−(1/2)) =Σ_(n=1) ^∞ (1/n)−Σ_(l=1) ^∞ (2/l)+Σ_(k=1) ^∞ (1/k)+2−1−(1/2) =Σ_(n=1) ^∞ ((1/n)−(2/n)+(1/n))+(1/2)=(1/2)
$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{{a}}{{n}}+\frac{{b}}{{n}+\mathrm{1}}+\frac{{c}}{{n}+\mathrm{2}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\:;\:{b}=−\mathrm{1}\:;\:{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{2}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}+\mathrm{1}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{2}} \\ $$$${let}\:\:{l}={n}+\mathrm{1}\:\:{and}\:\:{k}={n}+\mathrm{2} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{l}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{l}}+\underset{{k}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\left(\underset{{l}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{l}}−\frac{\mathrm{2}}{\mathrm{1}}\right)+\left(\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\underset{{l}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{l}}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}+\mathrm{2}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{2}}{{n}}+\frac{\mathrm{1}}{{n}}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *