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Erfan's question at Yahoo! Answers regarding a summation
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Hello Erfan,
We are given to evaluate:
\(\displaystyle S_n=\sum_{r=1}^n\left(\frac{r}{(2r-1)(2r+1)(2r+3)} \right)\)
We are instructed to use partial fraction decomposition on the summand, and so we assume it may be expressed in the form:
\(\displaystyle \frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{A}{2r-1}+\frac{B}{2r+1}+\frac{C}{2r+3}\)
Using the Heaviside cover-up method, we find:
\(\displaystyle A=\frac{\frac{1}{2}}{(2)(4)}=\frac{1}{16}\)
\(\displaystyle B=\frac{-\frac{1}{2}}{(-2)(2)}=\frac{1}{8}\)
\(\displaystyle C=\frac{-\frac{3}{2}}{(-4)(-2)}=-\frac{3}{16}\)
Hence:
\(\displaystyle \frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{16(2r-1)}+\frac{1}{8(2r+1)}-\frac{3}{16(2r+3)}\)
Factoring out \(\displaystyle \frac{1}{16}\), we may write the sum as follows:
\(\displaystyle S_n=\frac{1}{16}\left(\sum_{r=1}^n\left(\frac{1}{2r-1} \right)+2\sum_{r=1}^n\left(\frac{1}{2r+1} \right)-3\sum_{r=1}^n\left(\frac{1}{2r+3} \right) \right)\)
Re-indexing the sums, we have:
\(\displaystyle S_n=\frac{1}{16}\left(\sum_{r=1}^n\left(\frac{1}{2r-1} \right)+2\sum_{r=2}^{n+1}\left(\frac{1}{2r-1} \right)-3\sum_{r=3}^{n+2}\left(\frac{1}{2r-1} \right) \right)\)
Pulling off terms so that the indices of summation are the same for all three sums, we find:
\(\displaystyle S_n=\frac{1}{16}\left(\left(1+\frac{1}{3}+ \sum_{r=3}^n\left(\frac{1}{2r-1} \right) \right)+\right.\)
\(\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left.\left(\frac{2}{3}+2 \sum_{r=3}^n\left(\frac{1}{2r-1} \right)+\frac{2}{2n+1} \right)+\left(-3 \sum_{r=3}^n\left(\frac{1}{2r-1} \right)-\frac{3}{2n+1}-\frac{3}{2n+3} \right) \right)\)
Now the sums all add to zero, and we are left with:
\(\displaystyle S_n=\frac{1}{16}\left(2-\frac{1}{2n+1}-\frac{3}{2n+3} \right)=\frac{n(n+1)}{2(2n+1)(2n+3)}\)
We are given to evaluate:
\(\displaystyle S_n=\sum_{r=1}^n\left(\frac{r}{(2r-1)(2r+1)(2r+3)} \right)\)
We are instructed to use partial fraction decomposition on the summand, and so we assume it may be expressed in the form:
\(\displaystyle \frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{A}{2r-1}+\frac{B}{2r+1}+\frac{C}{2r+3}\)
Using the Heaviside cover-up method, we find:
\(\displaystyle A=\frac{\frac{1}{2}}{(2)(4)}=\frac{1}{16}\)
\(\displaystyle B=\frac{-\frac{1}{2}}{(-2)(2)}=\frac{1}{8}\)
\(\displaystyle C=\frac{-\frac{3}{2}}{(-4)(-2)}=-\frac{3}{16}\)
Hence:
\(\displaystyle \frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{16(2r-1)}+\frac{1}{8(2r+1)}-\frac{3}{16(2r+3)}\)
Factoring out \(\displaystyle \frac{1}{16}\), we may write the sum as follows:
\(\displaystyle S_n=\frac{1}{16}\left(\sum_{r=1}^n\left(\frac{1}{2r-1} \right)+2\sum_{r=1}^n\left(\frac{1}{2r+1} \right)-3\sum_{r=1}^n\left(\frac{1}{2r+3} \right) \right)\)
Re-indexing the sums, we have:
\(\displaystyle S_n=\frac{1}{16}\left(\sum_{r=1}^n\left(\frac{1}{2r-1} \right)+2\sum_{r=2}^{n+1}\left(\frac{1}{2r-1} \right)-3\sum_{r=3}^{n+2}\left(\frac{1}{2r-1} \right) \right)\)
Pulling off terms so that the indices of summation are the same for all three sums, we find:
\(\displaystyle S_n=\frac{1}{16}\left(\left(1+\frac{1}{3}+ \sum_{r=3}^n\left(\frac{1}{2r-1} \right) \right)+\right.\)
\(\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left.\left(\frac{2}{3}+2 \sum_{r=3}^n\left(\frac{1}{2r-1} \right)+\frac{2}{2n+1} \right)+\left(-3 \sum_{r=3}^n\left(\frac{1}{2r-1} \right)-\frac{3}{2n+1}-\frac{3}{2n+3} \right) \right)\)
Now the sums all add to zero, and we are left with:
\(\displaystyle S_n=\frac{1}{16}\left(2-\frac{1}{2n+1}-\frac{3}{2n+3} \right)=\frac{n(n+1)}{2(2n+1)(2n+3)}\)