# Sum of the series

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• Sep 7th 2012, 12:25 PM
alexander9408
Sum of the series
∑_(r=1)^n▒2/r(r+1)(r+2)
∑_(r=1)^n▒r/((2r-1)(2r+1)(2r+3))

I first convert both to partial fraction.
Then i use the method of differences to cancel off term by term, but i don't what term should i cancel.

Guide me through these two questions, please. Thank you.
• Sep 7th 2012, 01:20 PM
Plato
Re: Sum of the series
Quote:

Originally Posted by alexander9408
∑_(r=1)^n▒2/r(r+1)(r+2)
∑_(r=1)^n▒r/((2r-1)(2r+1)(2r+3))
I first convert both to partial fraction.
Then i use the method of differences to cancel off term by term, but i don't what term should i cancel.

$\displaystyle \sum\limits_{r = 1}^n {\frac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}} = \sum\limits_{r = 1}^n {\left[ {\frac{1}{r} - \frac{2}{{r + 1}} + \frac{1}{{r + 2}}} \right]}$

Now take $\displaystyle n=5$ write out all five terms and look for a pattern.
You may have to do that for several values.
• Sep 7th 2012, 07:02 PM
zaidalyafey
Re: Sum of the series
$\displaystyle \sum_{r=1}^{n}[(\frac{1}{r}-\frac{1}{r+1})+(\frac{1}{r+2}-\frac{1}{r+1})]$

$\displaystyle n=2\,\, \Rightarrow \,\, [(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{2})]+[(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{3})] =\frac{1}{2}-\frac{1}{3}+\frac{1}{4}$

$\displaystyle n=3\,\, \Rightarrow \,\, [(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{2})]+[(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{3})]+[(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{4})]=\frac{1}{2}-\frac{1}{4}+\frac{1}{5}$

$\displaystyle n=4\,\, \Rightarrow \,\, [(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{2})]+[(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{3})]+[(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{4})]+ [(\frac{1}{4}-\frac{1}{5})+(\frac{1}{6}-\frac{1}{5})]=\frac{1}{2}-\frac{1}{5}+\frac{1}{6}$

$\displaystyle \sum_{r=1}^{n}[(\frac{1}{r}-\frac{1}{r+1})+(\frac{1}{r+2}-\frac{1}{r+1})]=\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}$