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Alright so this series is infinite and arithmetic. $n=1$

$$\frac{1}{n(n+2)} = \frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \frac{1}{3\cdot5} + \cdots$$

I basically have no idea on how to solve this... any tips? But I know the answer should be $ \dfrac{3}{4}$.

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    $\begingroup$ Do you know how $$\sum_{n=1}^\infty \frac{1}{n(n+1)}$$ is determined? $\endgroup$
    – Daniel Fischer
    Feb 24, 2014 at 22:46
  • $\begingroup$ See telescoping series. $\endgroup$
    – Lucian
    Feb 24, 2014 at 22:48
  • $\begingroup$ it should be a telescopic series and should be done by partial fractions, if that helps? $\endgroup$
    – logicc
    Feb 24, 2014 at 22:50
  • $\begingroup$ @logicc, are you answering your own question? $\endgroup$
    – IAmNoOne
    Feb 24, 2014 at 23:04
  • $\begingroup$ It is not an arithmetic sequence. That would mean that the amount added to $1/(1\cdot3)$ to get $1/(2\cdot4)$ would be the same as the amound added to $1/(2\cdot4)$ to get $1/(3\cdot5)$. $\endgroup$
    – Michael Hardy
    Feb 25, 2014 at 1:39

2 Answers 2

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$$ \frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right) $$

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Hint: Decompose $\frac{1}{n(n+2)}$ into two fractions in order to get a telescoping series.

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