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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,



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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I found the question immediately online. How did you search? $\endgroup$ – Dietrich Burde Oct 7 '18 at 19:17
  • $\begingroup$ @DietrichBurde I first googled "summation 1/n(n+2)" but a lot of them autocorrected to n^2 so I couldn't find any results there. I then tried Symbolab and MathramAlpha because I didn't know what else to do. After that, I just adjusted my initial google with adding things like my summation parameters to try and make something appear on google. This question was very hard for me to find because I'm not really sure what to type in. $\endgroup$ – jjleahy Oct 7 '18 at 21:34

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 $${1\over k(k+2)} = {1\over 2}({1\over k}-{1\over k+2})$$


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