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How can I use the telescoping technique to compute the following sum? I'm having issues getting started. I know the basic steps but I don't know how to perform them. I know I have to separate the fraction into A and B. After that I have to perform the sum but I'm not sure what comes next.

$$\sum_{i=3}^n \frac{1}{i(i+3)} $$

Thanks for any help!

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    $\begingroup$ Could you show specifically where you're stuck? What'd you get after writing 1/i(i+3) as two fractions? $\endgroup$ – mathematician Apr 10 '14 at 5:06
  • $\begingroup$ Look instead at $a_i=\frac{3}{i(i+3)}$ write down the sum of the $a_i$ from $i=3$ to say $i=8$, and/or $i=9$, in this split form. We give a start: $\left(\frac{1}{3}-\frac{1}{6}\right)+ \left(\frac{1}{4}-\frac{1}{7}\right)+ \left(\frac{1}{5}-\frac{1}{8}\right)+\cdots$. (You will need more to see the cancellation pattern.) $\endgroup$ – André Nicolas Apr 10 '14 at 5:13
  • $\begingroup$ Will I end up with $1/n$ once the cancellation is over? $\endgroup$ – rubito Apr 10 '14 at 5:35
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$\sum_{i=3}^{n} \frac{1}{i(i+3)} = \frac{1}{3} \sum_{i=3}^{n} \frac{1}{i} - \frac{1}{i+3}$ via partial fractions.

To see what's going on, try writing out the first 4 or so terms and you'll quickly see what cancels out and what's left over...

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  • $\begingroup$ How did you go from the partial fractions $\frac1/{i(i+3)}$ to your sum? Shouldn't the expanded fraction be $\frac ({1/3}/{i})-({1-3}/{i+3}) $\endgroup$ – rubito Apr 10 '14 at 5:27
  • $\begingroup$ Nevermind, I figured that part out. So once I write out the first terms and cancel things what do I do? $\endgroup$ – rubito Apr 10 '14 at 5:29
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$\displaystyle\sum_{i=3}^n \dfrac{1}{i(i+3)}=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+3}\right)=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+1}+\dfrac{1}{i+1}-\dfrac{1}{i+2}+\dfrac{1}{i+2}-\dfrac{1}{i+3}\right)$

$\displaystyle=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+1}\right)+\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right)+\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i+2}-\dfrac{1}{i+3}\right)$

$=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{n+1}+\dfrac{1}{4}-\dfrac{1}{n+2}+\dfrac{1}{5}-\dfrac{1}{n+3}\right)$

$=\dfrac{(n-2)(47n^2+196n+189)}{180(n+1)(n+2)(n+3)}$

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  • $\begingroup$ Shouldn't it be: ${1/3}\sum_{i=3}^n {1/3}-{1/(n+3)}$ $\endgroup$ – rubito Apr 10 '14 at 6:19
  • $\begingroup$ @hxthanh. Which is $$\frac{47}{180}-\frac{1}{3 (n+1)}-\frac{1}{3 (n+2)}-\frac{1}{3 (n+3)}$$ $\endgroup$ – Claude Leibovici Apr 10 '14 at 6:22
  • $\begingroup$ So it's my answer right? I'm confused with all the extra steps that @hxthanh took. I just did the telescopic canceling and then substituted the $n$ term $\endgroup$ – rubito Apr 10 '14 at 6:28

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