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Question Number 44918 by Tawa1 last updated on 06/Oct/18
Find the sum to n terms of: (1/(1.2.3)) + (3/(2.3.4)) + (5/(3.4.5)) + ...
Findthesumtontermsof:11.2.3+32.3.4+53.4.5+…
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) .
letS=∑n=1∞2n−1n(n+1)(n+2)wehaveS=11.2.3+32.3.4+…S=limn→+∞SnwithSn=∑k=1n2k−1k(k+1)(k+2)letdecomposeF(x)=2x−1x(x+1)(x+2)F(x)=ax+bx+1+cx+2a=limx→0xF(x)=−12b=limx→−1(x+1)F(x)=−3(−1)(1)=3c=limx→−2(x+2)F(x)=−5(−2)(−1)=−52⇒F(x)=−12x+3x+1−52(x+2)⇒Sn=∑k=1nF(k)=−12∑k=1n1k+3∑k=1n1k+1−52∑k=1n1k+2but∑k=1n1k=Hn∑k=1n1k+1=∑k=2n+11k=Hn+1−1.∑k=1n1k+2=∑k=3n+21k=Hn+2−32⇒Sn=−12Hn+3Hn+1−3−52(Hn+2−32)=−12Hn+3Hn+1−52Hn+2+34butHn=ln(n)+γ+o(1n),Hn+1=ln(n+1)+γ+o(1n),Hn+2=ln(n+2)+γ+o(1n)SSn=−12ln(n)−γ2+3ln(n+1)+3γ−52ln(n+2)−52γ+o(1n)+34⇒Sn=ln(n+1)2−ln(n)−ln(n+2)52+34+o(1n)Sn=ln((n+1)2n(n+2)5)+34+o(1n)=ln(n2+2n+1(n+2)2n(n+2))+34+o(1n)butlimn→+∞ln(n2+2n+1(n+2)2n2+2n)=0⇒limn→+∞Sn=34⇒S=34.
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 .
thesumtontermsisSn=34−12Hn+3Hn+1−52Hn.
Commented by Tawa1 last updated on 06/Oct/18
God bless you sir. I appreciate. what is H sir
Godblessyousir.Iappreciate.whatisHsir
Commented by maxmathsup by imad last updated on 06/Oct/18
the harmonic serie H_n =Σ_(k=1) ^n (1/k) .
theharmonicserieHn=∑k=1n1k.
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)) + ... = ?
Sirplease,howtofindthesumofntermsof:11.2.3+14.5.6+17.8.9+…=?
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))) . _________________________
2S=[3−11.2.3+3(4−22.3.4)+5(5−33.4.5)+..+(2n−1)((n+2)−nn(n+1)(n+2))]=11.2−12.3+32.3−33.4+53.4−54.5+..+2n−1n(n+1)−2n−1(n+1)(n+2)]=(11.2+32.3+53.4+..+2n−1n(n+1))−(12.3+33.4+54.5+..+2n−1(n+1)(n+2))=[2−11.2+3(3−22.3)+5(4−33.4)+…+2n−1n−2n−1n+1]−[3−22.3+3(4−33.4)+5(5−44.5)+..+2n−1n+1−2n−1n+2]=[11−12+32−33+53−54+..+2n−1n−2n−1n+1]−[12−13+33−34+54−55+..+2n−3n−2n−3n+1+2n−1n+1−2n−1n+2]=32−(2−23)+(2−24)−….−2n+1n+1+2n−1n+2⇒2S=32−2n+1n+1+2n−1n+2S=34−2n+12(n+1)+2n−12(n+2)._________________________
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.
Godblessyousir.Butthequestionisfindthesuminnthterm.Answerintermsofn.Thanksforyouhelpsir.
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.
Pleaseviewagain,nowitsintermsofnandcorrecttoo.Nicequestion,thanksforsharing.
Commented by Tawa1 last updated on 06/Oct/18
Yes sir. God bless you sir. i really appreciate ..
Yessir.Godblessyousir.ireallyappreciate..
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)))
Tn=1+(n−1)2n(n+1)(n+2)Tn=2n−1n(n+1)(n+2)=an+bn+1+cn+22n−1=a(n+1)(n+2)+b(n)(n+2)+c(n)(n+1)putn=02a=−1a=−12n+1=0−3=b(−1)(1)b=3n+2=0−5=c(−2)(−1)c=−52Tn=−12(1n)+3(1n+1)+−52(1n+2)S=−12(11+12+13+14+…+1n)+3(12+13+14+…+1n+1)+−52(13+14+15+..+1n+2)letk=14+15+16+…+1nS=−12(1+12+13+k)+3(12+13+k+1n+1)−52(13+k+1n+1+1n+2)S=−12(116+k)+3(56+k+1n+1)−52(13+k+1n+1+1n+2)S=−1112+52−56+k(−12+3−52)+1n+1(3−52)+1n+2(−52)S=−11+30−1012+12(1n+1)−52(1n+2)S=34+12(1n+1)−12(1n+2)−2n+2S=34−2(1n+2)+12{n+2−n−1(n+2)(n+1)}S=34−2n+2+12(n+1)(n+2)
Commented by Tawa1 last updated on 06/Oct/18
God bless you sir. i appreciate your effort.
Godblessyousir.iappreciateyoureffort.
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.
Sir,itriedyourmethodhere:11.2.3+14.5.6+17.8.9+…Butidon′tgetanswer.

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