Find-the-sum-to-n-terms-of-1-1-2-3-3-2-3-4-5-3-4-5-

Question Number 44918 by Tawa1 last updated on 06/Oct/18

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{3}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\:+\:\frac{\mathrm{5}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}\:+\:…\: \\ $$

Commented by maxmathsup by imad last updated on 06/Oct/18

let S = Σ_(n=1) ^∞ ((2n−1)/(n(n+1)(n+2))) we have S =(1/(1.2.3)) +(3/(2.3.4)) +... S =lim_(n→+∞) S_n with S_n = Σ_(k=1) ^n ((2k−1)/(k(k+1)(k+2))) let decompose F(x) = ((2x−1)/(x(x+1)(x+2))) F(x)=(a/x) +(b/(x+1)) +(c/(x+2)) a=lim_(x→0) x F(x) = ((−1)/2) b =lim_(x→−1) (x+1)F(x) = ((−3)/((−1)(1))) =3 c =lim_(x→−2) (x+2)F(x) =((−5)/((−2)(−1))) =−(5/2) ⇒ F(x) =−(1/(2x)) +(3/(x+1)) −(5/(2(x+2))) ⇒ S_n =Σ_(k=1) ^n F(k) =−(1/2) Σ_(k=1) ^n (1/k) +3 Σ_(k=1) ^n (1/(k+1)) −(5/2) Σ_(k=1) ^n (1/(k+2)) but 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 =−(1/2) H_n +3H_(n+1) −3 −(5/2)( H_(n+2) −(3/2)) =−(1/2) H_n +3 H_(n+1) −(5/2) H_(n+2) +(3/4) but H_n =ln(n) +γ +o((1/n)) , H_(n+1) =ln(n+1) +γ +o((1/n)) ,H_(n+2) =ln (n+2)+γ +o((1/n))S S_n =−(1/2)ln(n)−(γ/2) +3ln(n+1) +3γ −(5/2)ln(n+2)−(5/2)γ +o((1/n))+(3/4) ⇒ S_n =ln(n+1)^2 −ln((√n))−ln(n+2)^(5/2) +(3/4) +o((1/n)) S_n =ln((((n+1)^2 )/( (√n)(√((n+2)^5 ))))) +(3/4) +o((1/n)) ^ =ln(((n^2 +2n+1)/((n+2)^2 (√(n(n+2)))))) +(3/4) +o((1/n)) but lim_(n→+∞) ln(((n^2 +2n+1)/((n+2)^2 (√(n^2 +2n)))))=0 ⇒lim_(n→+∞) S_n =(3/4) ⇒ S =(3/4) .

$${let}\:\:{S}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{2}{n}−\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:\:{we}\:{have}\:{S}\:=\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\frac{\mathrm{3}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\:+… \\ $$$${S}\:={lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} \:\:{with}\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{2}{k}−\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}\:\:{let}\:{decompose} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{2}{x}−\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{{x}+\mathrm{2}} \\ $$$${a}={lim}_{{x}\rightarrow\mathrm{0}} {x}\:{F}\left({x}\right)\:=\:\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${b}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)\:=\:\frac{−\mathrm{3}}{\left(−\mathrm{1}\right)\left(\mathrm{1}\right)}\:=\mathrm{3} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{2}} \left({x}+\mathrm{2}\right){F}\left({x}\right)\:=\frac{−\mathrm{5}}{\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)}\:=−\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}{x}}\:+\frac{\mathrm{3}}{{x}+\mathrm{1}}\:−\frac{\mathrm{5}}{\mathrm{2}\left({x}+\mathrm{2}\right)}\:\Rightarrow\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:+\mathrm{3}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:−\frac{\mathrm{5}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:\:{but}\: \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}+\mathrm{1}} −\mathrm{1}\:. \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:=\:\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}+\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\: \\ $$$${S}_{{n}} =−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:\:+\mathrm{3}{H}_{{n}+\mathrm{1}} −\mathrm{3}\:−\frac{\mathrm{5}}{\mathrm{2}}\left(\:{H}_{{n}+\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}} \:+\mathrm{3}\:{H}_{{n}+\mathrm{1}} \:−\frac{\mathrm{5}}{\mathrm{2}}\:{H}_{{n}+\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\:\:\:{but} \\ $$$${H}_{{n}} ={ln}\left({n}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\:\:,\:\:{H}_{{n}+\mathrm{1}} ={ln}\left({n}+\mathrm{1}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:,{H}_{{n}+\mathrm{2}} ={ln}\:\left({n}+\mathrm{2}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right){S} \\ $$$${S}_{{n}} =−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)−\frac{\gamma}{\mathrm{2}}\:\:+\mathrm{3}{ln}\left({n}+\mathrm{1}\right)\:+\mathrm{3}\gamma\:\:−\frac{\mathrm{5}}{\mathrm{2}}{ln}\left({n}+\mathrm{2}\right)−\frac{\mathrm{5}}{\mathrm{2}}\gamma\:\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)+\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow \\ $$$${S}_{{n}} ={ln}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \:−{ln}\left(\sqrt{{n}}\right)−{ln}\left({n}+\mathrm{2}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:\:+\frac{\mathrm{3}}{\mathrm{4}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$${S}_{{n}} ={ln}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\:\sqrt{{n}}\sqrt{\left({n}+\mathrm{2}\right)^{\mathrm{5}} }}\right)\:+\frac{\mathrm{3}}{\mathrm{4}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\overset{} {\:}={ln}\left(\frac{{n}^{\mathrm{2}} \:+\mathrm{2}{n}+\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{{n}\left({n}+\mathrm{2}\right)}}\right)\:+\frac{\mathrm{3}}{\mathrm{4}}\:+{o}\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$${but}\:{lim}_{{n}\rightarrow+\infty} \:{ln}\left(\frac{{n}^{\mathrm{2}} \:+\mathrm{2}{n}+\mathrm{1}}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} \sqrt{{n}^{\mathrm{2}} \:+\mathrm{2}{n}}}\right)=\mathrm{0}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{3}}{\mathrm{4}}\:. \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 06/Oct/18

the sum to n terms is S_n =(3/4) −(1/2)H_n +3H_(n+1) −(5/2) H_n .

$${the}\:{sum}\:{to}\:{n}\:{terms}\:{is}\:{S}_{{n}} =\frac{\mathrm{3}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} \:+\mathrm{3}{H}_{{n}+\mathrm{1}} −\frac{\mathrm{5}}{\mathrm{2}}\:{H}_{{n}} . \\ $$

Commented by Tawa1 last updated on 06/Oct/18

God bless you sir. I appreciate. what is H sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{appreciate}. \\ $$$$\mathrm{what}\:\mathrm{is}\:\boldsymbol{\mathrm{H}}\:\mathrm{sir} \\ $$

Commented by maxmathsup by imad last updated on 06/Oct/18

the harmonic serie H_n =Σ_(k=1) ^n (1/k) .

$${the}\:{harmonic}\:{serie}\:{H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:. \\ $$

Commented by Tawa1 last updated on 09/Oct/18

Sir please, how to find the sum of n terms of: (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ... = ?

$$\mathrm{Sir}\:\mathrm{please},\:\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:+\:…\:\:=\:\:? \\ $$

Answered by ajfour last updated on 06/Oct/18

2S = [((3−1)/(1.2.3))+3(((4−2)/(2.3.4)))+5(((5−3)/(3.4.5)))+..+(2n−1)((((n+2)−n)/(n(n+1)(n+2))))] = (1/(1.2))−(1/(2.3))+(3/(2.3))−(3/(3.4))+(5/(3.4))−(5/(4.5))+..+((2n−1)/(n(n+1)))−((2n−1)/((n+1)(n+2)))] = ((1/(1.2))+(3/(2.3))+(5/(3.4))+..+((2n−1)/(n(n+1))))−((1/(2.3))+(3/(3.4))+(5/(4.5))+..+((2n−1)/((n+1)(n+2)))) = [((2−1)/(1.2))+3(((3−2)/(2.3)))+5(((4−3)/(3.4)))+...+((2n−1)/n)−((2n−1)/(n+1))] −[((3−2)/(2.3))+3(((4−3)/(3.4)))+5(((5−4)/(4.5)))+..+((2n−1)/(n+1))−((2n−1)/(n+2))] = [(1/1)−(1/2)+(3/2)−(3/3)+(5/3)−(5/4)+..+((2n−1)/n)−((2n−1)/(n+1))] −[(1/2)−(1/3)+(3/3)−(3/4)+(5/4)−(5/5)+..+((2n−3)/n)−((2n−3)/(n+1))+((2n−1)/(n+1))−((2n−1)/(n+2))] = (3/2)−(((2−2)/3))+(((2−2)/4))−....−((2n+1)/(n+1))+((2n−1)/(n+2)) ⇒ 2S = (3/2)−((2n+1)/(n+1))+((2n−1)/(n+2)) S = (3/4) −((2n+1)/(2(n+1)))+((2n−1)/(2(n+2))) . _________________________

$$\:\mathrm{2}{S}\:=\:\left[\frac{\mathrm{3}−\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}+\mathrm{3}\left(\frac{\mathrm{4}−\mathrm{2}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\right)+\mathrm{5}\left(\frac{\mathrm{5}−\mathrm{3}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}\right)+..+\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\frac{\left({n}+\mathrm{2}\right)−{n}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right)\right] \\ $$$$\left.\:=\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}.\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{3}.\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{3}.\mathrm{4}}−\frac{\mathrm{5}}{\mathrm{4}.\mathrm{5}}+..+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{2}{n}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right] \\ $$$$\:=\:\left(\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}.\mathrm{3}}+\frac{\mathrm{5}}{\mathrm{3}.\mathrm{4}}+..+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\right)−\left(\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{3}.\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{4}.\mathrm{5}}+..+\frac{\mathrm{2}{n}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right) \\ $$$$\:=\:\left[\frac{\mathrm{2}−\mathrm{1}}{\mathrm{1}.\mathrm{2}}+\mathrm{3}\left(\frac{\mathrm{3}−\mathrm{2}}{\mathrm{2}.\mathrm{3}}\right)+\mathrm{5}\left(\frac{\mathrm{4}−\mathrm{3}}{\mathrm{3}.\mathrm{4}}\right)+…+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}−\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{1}}\right] \\ $$$$\:\:\:\:\:\:−\left[\frac{\mathrm{3}−\mathrm{2}}{\mathrm{2}.\mathrm{3}}+\mathrm{3}\left(\frac{\mathrm{4}−\mathrm{3}}{\mathrm{3}.\mathrm{4}}\right)+\mathrm{5}\left(\frac{\mathrm{5}−\mathrm{4}}{\mathrm{4}.\mathrm{5}}\right)+..+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{2}}\right] \\ $$$$\:=\:\left[\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{3}}+\frac{\mathrm{5}}{\mathrm{3}}−\frac{\mathrm{5}}{\mathrm{4}}+..+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}−\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{1}}\right] \\ $$$$\:\:\:\:\:−\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{4}}−\frac{\mathrm{5}}{\mathrm{5}}+..+\frac{\mathrm{2}{n}−\mathrm{3}}{{n}}−\frac{\mathrm{2}{n}−\mathrm{3}}{{n}+\mathrm{1}}+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{2}}\right] \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}−\left(\frac{\mathrm{2}−\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{2}−\mathrm{2}}{\mathrm{4}}\right)−….−\frac{\mathrm{2}{n}+\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{S}\:=\:\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{2}{n}+\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{2}{n}−\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\:\:\:{S}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:−\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{2}\right)}\:. \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\: \\ $$

Commented by Tawa1 last updated on 06/Oct/18

God bless you sir. But the question is find the sum in nth term. Answer in terms of n. Thanks for you help sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{But}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{in}\:\mathrm{nth}\:\mathrm{term}.\:\:\mathrm{Answer}\:\mathrm{in}\:\mathrm{terms}\: \\ $$$$\mathrm{of}\:\mathrm{n}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{you}\:\mathrm{help}\:\mathrm{sir}. \\ $$

Commented by ajfour last updated on 06/Oct/18

$${Please}\:{view}\:{again},\:{now}\:{its}\:{in} \\ $$$${terms}\:{of}\:{n}\:{and}\:{correct}\:{too}. \\ $$$${Nice}\:{question},\:{thanks}\:{for}\:{sharing}. \\ $$

Commented by Tawa1 last updated on 06/Oct/18

Yes sir. God bless you sir. i really appreciate ..

$$\mathrm{Yes}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:..\: \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Oct/18

T_n =((1+(n−1)2)/(n(n+1)(n+2))) T_n =((2n−1)/(n(n+1)(n+2)))=(a/n)+(b/(n+1))+(c/(n+2)) 2n−1=a(n+1)(n+2)+b(n)(n+2)+c(n)(n+1) put n=0 2a=−1 a=((−1)/2) n+1=0 −3=b(−1)(1) b=3 n+2=0 −5=c(−2)(−1) c=((−5)/2) T_n =((−1)/2)((1/n))+3((1/(n+1)))+((−5)/2)((1/(n+2))) S=((−1)/2)((1/1)+(1/2)+(1/3)+(1/4)+...+(1/n))+ 3((1/2)+(1/3)+(1/4)+...+(1/(n+1)))+ ((−5)/2)((1/3)+(1/4)+(1/5)+..+(1/(n+2))) let k=(1/4)+(1/5)+(1/6)+...+(1/n) S=((−1)/2)(1+(1/2)+(1/3)+k)+3((1/2)+(1/3)+k+(1/(n+1))) −(5/2)((1/3)+k+(1/(n+1))+(1/(n+2))) S=((−1)/2)(((11)/6)+k)+3((5/6)+k+(1/(n+1)))−(5/2)((1/3)+k+(1/(n+1))+(1/(n+2))) S=((−11)/(12))+(5/2)−(5/6)+k(((−1)/2)+3−(5/2))+(1/(n+1))(3−(5/2))+(1/(n+2))(((−5)/2)) S=((−11+30−10)/(12))+(1/2)((1/(n+1)))−(5/2)((1/(n+2))) S=(3/4)+(1/2)((1/(n+1)))−(1/2)((1/(n+2)))−(2/(n+2)) S=(3/4)−2((1/(n+2)))+(1/2){((n+2−n−1)/((n+2)(n+1)))} S=(3/4)−(2/(n+2))+(1/(2(n+1)(n+2)))

$${T}_{{n}} =\frac{\mathrm{1}+\left({n}−\mathrm{1}\right)\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$${T}_{{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{{a}}{{n}}+\frac{{b}}{{n}+\mathrm{1}}+\frac{{c}}{{n}+\mathrm{2}} \\ $$$$\mathrm{2}{n}−\mathrm{1}={a}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)+{b}\left({n}\right)\left({n}+\mathrm{2}\right)+{c}\left({n}\right)\left({n}+\mathrm{1}\right) \\ $$$${put}\:{n}=\mathrm{0} \\ $$$$\mathrm{2}{a}=−\mathrm{1}\:\:{a}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${n}+\mathrm{1}=\mathrm{0} \\ $$$$−\mathrm{3}={b}\left(−\mathrm{1}\right)\left(\mathrm{1}\right)\:\:\:{b}=\mathrm{3} \\ $$$${n}+\mathrm{2}=\mathrm{0} \\ $$$$−\mathrm{5}={c}\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)\:\:\:\:{c}=\frac{−\mathrm{5}}{\mathrm{2}} \\ $$$${T}_{{n}} =\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}}\right)+\mathrm{3}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)+\frac{−\mathrm{5}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$${S}=\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{n}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\frac{−\mathrm{5}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+..+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$${let}\:{k}=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+…+\frac{\mathrm{1}}{{n}} \\ $$$${S}=\frac{−\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+{k}\right)+\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+{k}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$\:\:\:\:\:\:\:\:−\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}+{k}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$${S}=\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{11}}{\mathrm{6}}+{k}\right)+\mathrm{3}\left(\frac{\mathrm{5}}{\mathrm{6}}+{k}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}+{k}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$${S}=\frac{−\mathrm{11}}{\mathrm{12}}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{6}}+{k}\left(\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{3}−\frac{\mathrm{5}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\mathrm{3}−\frac{\mathrm{5}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{{n}+\mathrm{2}}\left(\frac{−\mathrm{5}}{\mathrm{2}}\right) \\ $$$${S}=\frac{−\mathrm{11}+\mathrm{30}−\mathrm{10}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\frac{\mathrm{5}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$${S}=\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)−\frac{\mathrm{2}}{{n}+\mathrm{2}} \\ $$$${S}=\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{2}\left(\frac{\mathrm{1}}{{n}+\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{{n}+\mathrm{2}−{n}−\mathrm{1}}{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}\right\} \\ $$$${S}=\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{2}}{{n}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$ \\ $$

Commented by Tawa1 last updated on 06/Oct/18

God bless you sir. i appreciate your effort.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}.\: \\ $$

Commented by Tawa1 last updated on 07/Oct/18

Sir, i tried your method here: (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ... But i don′t get answer.

$$\mathrm{Sir},\:\mathrm{i}\:\mathrm{tried}\:\mathrm{your}\:\mathrm{method}\:\mathrm{here}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:+\:… \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{answer}.\:\: \\ $$

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