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n-1-1-n-n-1-




Question Number 93859 by  M±th+et+s last updated on 15/May/20
Σ_(n=1) ^∞  (1/(n(n+1)))
n=11n(n+1)
Answered by Ar Brandon last updated on 15/May/20
(1/(n(n+1)))=(1/n)−(1/(n+1))⇒Σ_(n=1) ^∞ (1/(n(n+1)))=Σ_(n=1) ^∞ [(1/n)−(1/(n+1))]=lim_(k→+∞) Σ_(n=1) ^k [(1/n)−(1/(n+1))]  =lim_(k→+∞) [1−(1/2)+(1/2)−(1/3)+(1/3)−(1/4)+∙∙∙+(1/k)−(1/(k+1))]  =lim_(k→+∞) [1−(1/(k+1))]=1
1n(n+1)=1n1n+1n=11n(n+1)=n=1[1n1n+1]=limk+kn=1[1n1n+1]=limk+[112+1213+1314++1k1k+1]=limk+[11k+1]=1

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