Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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}$.

share|cite|improve this question
Do you know how $$\sum_{n=1}^\infty \frac{1}{n(n+1)}$$ is determined? – Daniel Fischer Feb 24 '14 at 22:46
See telescoping series. – Lucian Feb 24 '14 at 22:48
it should be a telescopic series and should be done by partial fractions, if that helps? – logicc Feb 24 '14 at 22:50
@logicc, are you answering your own question? – Nameless Feb 24 '14 at 23:04
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 $1/(3\cdot5)$. – Michael Hardy Feb 25 '14 at 1:39

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

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.