Help with Telescopic Series with 3 terms in denominator

All the examples i have done and seen only have 2 terms in the denominator so I am a bit stuck with this one. I have attached what I have done so far, not sure how to proceed with it.

Thank you for the hints they were useful, after working it out more I ended up with the following but now i am confused on what to do next, do I have to do another partial fraction decomposition? My work after the hints

• Note that our expression is equal to $\frac{1/2}{n(n+1)}-\frac{1/2}{(n+1)(n+2)}$. – André Nicolas Jul 20 '16 at 4:05
• About the added work: The limit of $\frac{1}{4}-\frac{1/2}{(n+1)(n+2)}$ is $1/4$. – André Nicolas Jul 20 '16 at 5:57
• so the limit of the second fraction equals to 0? from what I know it would be .5/ (infinity*infinity) right? Also thanks for all your help! summer calculus class is very fast which makes it difficult! – Jessey malow Jul 20 '16 at 6:52
• I don't like to treat "$\infty$" as a number. Imagine that $n$ grows without bound. Then $\frac{1/2}{(n+1)(n+2)}$ approaches $0$. Much more informally, perhaps too informally, all the terms except the first get cancelled out. – André Nicolas Jul 20 '16 at 6:57

You are close to the end. Express what you got as $$\left(\frac{1/2}{n}-\frac{1/2}{n+1}\right)-\left(\frac{1/2}{n+1}-\frac{1/2}{n+2}\right).$$ It looks a little better as $$\frac{1/2}{n(n+1)}-\frac{1/2}{(n+1)(n+2)}.$$ Now add up, and (in either version) watch almost all the terms cancel.
Hint: \begin{align} \sum_{n=1}^\infty\frac1{n(n+1)(n+2)} &=\frac12\sum_{n=1}^\infty\left[\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right]\\ &=\frac12\sum_{n=1}^\infty\frac1{n(n+1)}-\frac12\sum_{n=2}^\infty\frac1{n(n+1)}\\ \end{align} More Generally \begin{align} &\frac1{n(n+1)\dots(n+k-1)}-\frac1{(n+1)(n+2)\dots(n+k)}\\ &=\frac1{(n+1)(n+2)\dots(n+k-1)}\left(\frac1n-\frac1{n+k}\right)\\ &=\frac{k}{n(n+1)\dots(n+k)} \end{align} Therefore, $$\frac1{n(n+1)\dots(n+k)}=\frac1k\left[\frac1{n(n+1)\dots(n+k-1)}-\frac1{(n+1)(n+2)\dots(n+k)}\right]$$