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

The series is,
$\sum _{n=9}^{\mathrm{\infty }}\frac{1}{n\left(n-1\right)}$
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Toni Scott
$\sum _{n=9}^{\mathrm{\infty }}\frac{1}{n\left(n-1\right)}=\sum _{n=9}^{\mathrm{\infty }}\left(\frac{1}{n-1}-\frac{1}{n}\right)$
$=\left(\frac{1}{8}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+\dots$
$=\frac{1}{9}+\left(-\frac{1}{9}+\frac{1}{9}\right)+\left(-\frac{1}{10}+\frac{1}{10}\right)+\dots$
$=\frac{1}{8}$
###### Not exactly what you’re looking for?
Fasaniu

the general method:
$\frac{1}{n\left(n-1\right)}=\frac{A}{n}+\frac{B}{n-1}$
$B=lim\left(n-1\right)×\frac{1}{n\left(n-1\right)}=-1$
$A=limn×\frac{1}{n\left(n-1\right)}=-1$
This is why you get
$\frac{1}{n\left(n-1\right)}=\frac{1}{n-1}-\frac{1}{n}$
and then (keep the first term for the lower index and the last with the bigger)
${S}_{N}=\sum _{n=9}^{N}\frac{1}{n-1}-\frac{1}{n}=\frac{1}{8}-\frac{1}{N}$

###### Not exactly what you’re looking for?
user_27qwe

$\frac{1}{n\left(n-1\right)}=\frac{A}{n}+\frac{B}{n-1}=\frac{A\left(n-1\right)+Bn}{n\left(n-1\right)}=\frac{\left(A+B\right)n-A}{n\left(n+1\right)}$
$A+B=0\to A=-B$

$\frac{1}{n\left(n-1\right)}=\frac{1}{n-1}-\frac{1}{n}$
Now write out some terms:
$\left(\frac{1}{8}-\frac{1}{9}\right)+\left(\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+...$
It's clear that the $-\frac{1}{n}$ is canceling the next term. So it should be clear that this gives:
$\sum _{9}^{N}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{8}-\frac{1}{N}$
Now take the limit as N goers to infinity:
$\underset{N\to \mathrm{\infty }}{lim}\sum _{9}^{N}\left(\frac{1}{n-1}-\frac{1}{N}\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{8}-\frac{1}{N}=\frac{1}{8}$