Summation if series

**alexander9408** · Sep 10th 2012, 06:54 AM

Hello there, could anyone help me to solve this one?
∑_(r=1)^n▒r/((2r-1)(2r+1)(2r+3)) , I already changed it to the partial fraction

So what's next? Any help with detail explanation will greatly appreciate.Thanks.

**Soroban** · Sep 10th 2012, 09:52 AM

Hello, alexander9408!

Hello there, could anyone help me to solve this one?

. . $\displaystyle S \;=\; \sum^n_{r=1} \frac{r}{(2r-1)(2r+1)(2r+3)}$

I already changed it to the partial fraction : . $\frac{1}{16(2r-1)} + \frac{1}{8(2r+1)} - \frac{3}{16(2r+3)}$
For clarity, write the denominators in increasing or decreasing order.

So what's next?

$\text{We have: }\:S \;=\;\frac{1}{16}\sum^n_{r=1}\left(\frac{1}{2r-1} + \frac{2}{2r+1} - \frac{3}{2r+3}\right)$

Now we let $r \:=\:1,2,3\:\hdots\: n$
. . I will omit the leading fraction for now.

$\text{We have: }\:\left(\tfrac{1}{1} - \tfrac{2}{3} - \tfrac{3}{5}\right) + \left(\tfrac{1}{3} + \tfrac{2}{5} - \tfrac{3}{7}\right) + \left(\tfrac{1}{5} + \tfrac{2}{7} - \tfrac{3}{9}\right) + \cdots$

. . . . . . . . . . . $+ \left(\tfrac{1}{2n-1} + \tfrac{2}{2n+1} - \tfrac{3}{2n+3}\right) + \left(\tfrac{2}{2n+1} + \tfrac{2}{2n+3} - \tfrac{3}{2n+5}\right)$

"Collect" the fractions by their denominators:

. . $1 + \left(\tfrac{2}{3} + \tfrac{1}{3}\right) + \left(-\tfrac{3}{5} + \tfrac{2}{5} + \tfrac{1}{5}\right) + \left(-\tfrac{3}{7} + \tfrac{2}{7} + \tfrac{1}{7}\right) + \cdots$

. . . . . . . $+ \left(-\tfrac{3}{2n+1} + \tfrac{2}{2n+1} + \tfrac{1}{2n+1}\right) + \left(-\tfrac{3}{2n+3} + \tfrac{2}{2n+3}\right) - \tfrac{3}{2n+5}$

Most of the terms cancel. .All that remains is:

. . $1 + 1 - \frac{1}{2n+3} - \frac{3}{2n+5} \;=\;\frac{8n^2+24n+16)}{(2n+3)(2n+5)} \;=\;\frac{8(n+3n+2)}{(2n+3)(2n+5)}$

$\text{Therefore: }\:S \;=\;\frac{1}{16}\cdot\frac{8(n+1)(n+2)}{(2n+3)(2n +5)} \;=\;\frac{(n+1)(n+2)}{2(2n+3)(2n+5)}$

But check my work . . . please!

**Plato** · Sep 10th 2012, 02:16 PM

Originally Posted by Soroban

Hello, alexander9408!

$\text{We have: }\:S \;=\;\frac{1}{16}\sum^n_{r=1}\left(\frac{1}{2r-1} + \frac{2}{2r+1} - \frac{3}{2r+3}\right)$

But check my work . . please!

I would write
$\sum^n_{r=1}\left(\frac{1}{2r-1} + \frac{2}{2r+1} - \frac{3}{2r+3}\right)=\sum^n_{r=1}\left(\frac{1}{2 r-1} - \frac{1}{2r+1} \right)+3\sum^n_{r=1}\left( \frac{1}{2r+1} - \frac{1}{2r+3}\right)$ .

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