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

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

compagnia04 2021-12-25 Answered

The series is,

\sum_{n = 9}^{\infty} \frac{1}{n (n - 1)}

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Toni Scott

Answered 2021-12-26 Author has 32 answers

\sum_{n = 9}^{\infty} \frac{1}{n (n - 1)} = \sum_{n = 9}^{\infty} (\frac{1}{n - 1} - \frac{1}{n})

= (\frac{1}{8} - \frac{1}{9}) + (\frac{1}{9} - \frac{1}{10}) + (\frac{1}{10} - \frac{1}{11}) + \dots

= \frac{1}{9} + (- \frac{1}{9} + \frac{1}{9}) + (- \frac{1}{10} + \frac{1}{10}) + \dots

= \frac{1}{8}

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Fasaniu

Answered 2021-12-27 Author has 46 answers

the general method:
$\frac{1}{n (n - 1)} = \frac{A}{n} + \frac{B}{n - 1}$
$B = lim (n - 1) \times \frac{1}{n (n - 1)} = - 1$
$A = lim n \times \frac{1}{n (n - 1)} = - 1$
This is why you get
$\frac{1}{n (n - 1)} = \frac{1}{n - 1} - \frac{1}{n}$
and then (keep the first term for the lower index and the last with the bigger)
$S_{N} = \sum_{n = 9}^{N} \frac{1}{n - 1} - \frac{1}{n} = \frac{1}{8} - \frac{1}{N}$

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user_27qwe

Answered 2021-12-30 Author has 208 answers

$\frac{1}{n (n - 1)} = \frac{A}{n} + \frac{B}{n - 1} = \frac{A (n - 1) + B n}{n (n - 1)} = \frac{(A + B) n - A}{n (n + 1)}$
$A + B = 0 \to A = - B$
$- A = 1 \to A = - 1, B = 1$
$\frac{1}{n (n - 1)} = \frac{1}{n - 1} - \frac{1}{n}$
Now write out some terms:
$(\frac{1}{8} - \frac{1}{9}) + (\frac{1}{9}) + (\frac{1}{9} - \frac{1}{10}) + (\frac{1}{10} - \frac{1}{11}) + . . .$
It's clear that the $- \frac{1}{n}$ is canceling the next term. So it should be clear that this gives:
$\sum_{9}^{N} (\frac{1}{n - 1} - \frac{1}{n}) = \frac{1}{8} - \frac{1}{N}$
Now take the limit as N goers to infinity:
$lim_{N \to \infty} \sum_{9}^{N} (\frac{1}{n - 1} - \frac{1}{N}) = lim_{n \to \infty} \frac{1}{8} - \frac{1}{N} = \frac{1}{8}$

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