# Fraction Sum Series

$1/2 + 1/6 + 1/12+ 1/20 + 1/30 + 1/42 +1/56 + 1/72 + 1/90 + 1/110 +1/132$

I have noticed that it can be written as $1/(1*2) + 1/(2*3) +1/(3*4) + 1/(4*5).... +1/(11*12)$

However I don't know how to continue..

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Parentheses, please. $1/1*2$ can be read as $(1/1)*2$ which is clearly not what you mean. – Ross Millikan Feb 13 at 2:35

Hint:

Your sum can be written in the following form:

$$\sum\limits_{n=1}^{11} \dfrac{1}{n(n+1)}$$

Apply partial fraction decomposition to $\frac{1}{n(n+1)}$ to write it in the form $\frac{A}{n}+\frac{B}{n+1}$ and see what happens as you simplify.

The series should telescope. I.e. adjacent terms will cancel with one another, leaving you with only the first and last terms not cancelled. You have $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. The series then looks like $\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\cdots-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}$

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Then note that $\frac 1{1\cdot 2}$ = $\frac 11 - \frac 12$ and the sum telescopes.

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