Menu Close

n-1-n-n-2-n-1-

Tinku Tara 0 Comments
Pin




Question Number 123167 by Dwaipayan Shikari last updated on 23/Nov/20
Σ_(n=1) ^∞ ((√n)/(n^2 +n+1))
∑∞n=1nn2+n+1
Commented by mathmax by abdo last updated on 23/Nov/20
Σ_(n=1) ^∞ ((√n)/(n^2 +n+1))=((√1)/3) +((√2)/(2^2 +2+1)) +((√3)/(3^2 +3+1))+.....
∑n=1∞nn2+n+1=13+222+2+1+332+3+1+…..
Answered by liberty last updated on 23/Nov/20
Σ_(n=1) ^∞ ((√n)/((n+1+(√n))(n+1−(√n)))) Σ_(t=1) ^∞ (t/((t^2 +t+1)(t^2 −t+1))) (1/2)Σ_(t=1) ^∞ (((t^2 +t+1)−(t^2 −t+1))/((t^2 +t+1)(t^2 −t+1))) (1/2)[ Σ_(t=1) ^∞ (1/(t^2 −t+1))−(1/(t^2 +t+1)) ] (1/2) [ 1−(1/3)+(1/3)−(1/7)+(1/7)−(1/(13))+(1/(13))−(1/(21))+... +0−0 ] (1/2)×1=(1/2)
∑∞n=1n(n+1+n)(n+1−n)∑∞t=1t(t2+t+1)(t2−t+1)12∑∞t=1(t2+t+1)−(t2−t+1)(t2+t+1)(t2−t+1)12[∑∞t=11t2−t+1−1t2+t+1]12[1−13+13−17+17−113+113−121+…+0−0]12×1=12
Commented by Dwaipayan Shikari last updated on 23/Nov/20
But ′t′ doesn′t sum up linearly (1/2)Σ_(n=1) ^∞ (1/((n+1−(√n))))−(1/((n+1+(√n)))) =(1/2)Σ_(t=(√1)) ^∞ (1/(t^2 −t+1))−(1/(t^2 +t+1)) ′t′ will take take the value of (√1), (√2), (√3), ....
But′t′doesn′tsumuplinearly12∑∞n=11(n+1−n)−1(n+1+n)=12∑∞t=11t2−t+1−1t2+t+1′t′willtaketakethevalueof1,2,3,….
Commented by mathmax by abdo last updated on 23/Nov/20
answer not correct you can not use (√n)=t in sum its not a integral ....!
answernotcorrectyoucannotusen=tinsumitsnotaintegral….!
Commented by benjo_mathlover last updated on 23/Nov/20
it should be Σ_(n=1) ^∞ ((√n)/((n+(√n)+1)(n−(√n)+1)))= Σ_(n=1) ^∞ (((n+(√n)+1)−(n−(√n)+1))/(2(n+(√n)+1)(n−(√n)+1)))= (1/2)Σ_(n=1) ^∞ (1/(n−(√n)+1)) − (1/(n+(√n)+1))= (1/2) [ 1−(1/3) + (1/(3−(√2)))−(1/(3+(√2)))+(1/(4−(√3)))−(1/(4+(√3)))+... ]
itshouldbe∑∞n=1n(n+n+1)(n−n+1)=∑∞n=1(n+n+1)−(n−n+1)2(n+n+1)(n−n+1)=12∑∞n=11n−n+1−1n+n+1=12[1−13+13−2−13+2+14−3−14+3+…]

Pin

Leave a Reply

Your email address will not be published.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