# $\sum_{k=1}^{519}\frac{1}{k(k+2)}$, how do I solve this? [duplicate]

$$\sum_{k=1}^{519}\frac{1}{k(k+2)}$$

I've been trying to figure out this summation for a while now but I can't seem to get it. I've been looking at my notes and the only trick I've been given for summations like this one is $$\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$$. This doesn't help me to my knowledge because this kind of fraction decomposition doesn't work on my question. I have premium Symbolab and that says it doesn't support this kind of question. WolframAlpha gives me the decimal approximation of 0.7480 but no support work.

I'm sorry if this question has been posted here before but I couldn't find this question anywhere online.

Thank you for the help,

-jjleahy

## marked as duplicate by Dietrich Burde, Parcly Taxel, Nosrati, Key Flex, Don ThousandOct 8 '18 at 2:33

hint: Write $$\dfrac{1}{k(k+2)} = \dfrac{1}{2}\left(\dfrac{1}{k} - \dfrac{1}{k+1}\right) + \dfrac{1}{2}\left(\dfrac{1}{k+1} - \dfrac{1}{k+2}\right)$$ . Telescoping is underway...
Hint: Not that $${1\over k(k+2)} = {1\over 2}({1\over k}-{1\over k+2})$$