The series is, \sum_{n=9}^\infty \frac{1}{n(n-1)}

compagnia04 2021-12-25 Answered
The series is,
n=91n(n1)
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Expert Answer

Toni Scott
Answered 2021-12-26 Author has 32 answers
n=91n(n1)=n=9(1n11n)
=(1819)+(19110)+(110111)+
=19+(19+19)+(110+110)+
=18
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Fasaniu
Answered 2021-12-27 Author has 46 answers

the general method:
1n(n1)=An+Bn1
B=lim(n1)×1n(n1)=1
A=limn×1n(n1)=1
This is why you get
1n(n1)=1n11n
and then (keep the first term for the lower index and the last with the bigger)
SN=n=9N1n11n=181N

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user_27qwe
Answered 2021-12-30 Author has 208 answers

1n(n1)=An+Bn1=A(n1)+Bnn(n1)=(A+B)nAn(n+1)
A+B=0A=B
A=1A=1, B=1
1n(n1)=1n11n
Now write out some terms:
(1819)+(19)+(19110)+(110111)+...
It's clear that the 1n is canceling the next term. So it should be clear that this gives:
9N(1n11n)=181N
Now take the limit as N goers to infinity:
limN9N(1n11N)=limn181N=18

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