2
$\begingroup$

enter image description here

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

$\endgroup$
  • $\begingroup$ Note that our expression is equal to $\frac{1/2}{n(n+1)}-\frac{1/2}{(n+1)(n+2)}$. $\endgroup$ – André Nicolas Jul 20 '16 at 4:05
  • $\begingroup$ About the added work: The limit of $\frac{1}{4}-\frac{1/2}{(n+1)(n+2)}$ is $1/4$. $\endgroup$ – André Nicolas Jul 20 '16 at 5:57
  • $\begingroup$ 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! $\endgroup$ – Jessey malow Jul 20 '16 at 6:52
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Jul 20 '16 at 6:57
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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, $$ \frac1{n(n+1)\dots(n+k)}=\frac1k\left[\frac1{n(n+1)\dots(n+k-1)}-\frac1{(n+1)(n+2)\dots(n+k)}\right] $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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