Answer: (i) Expanding the right-hand side gives
\(\frac{1}{2}\left\{\frac{(2r+1)(2r+3)}{(r+1)(r+2)}-\frac{(2r-1)(2r+1)}{r(r+1)}\right\}\)
\(=\frac{1}{2}\left\{\frac{(2r+1)(2r+3)r-(2r-1)(2r+1)(r+2)}{r(r+1)(r+2)}\right\}.\)
Now
\((2r+1)(2r+3)r= r(4r^2+8r+3)=4r^3+8r^2+3r,\)
and
\((2r-1)(2r+1)(r+2)=(4r^2-1)(r+2)=4r^3+8r^2-r-2.\)
So the numerator becomes
\(4r^3+8r^2+3r-(4r^3+8r^2-r-2)=4r+2.\)
Hence the right-hand side is
\(\frac{1}{2}\cdot\frac{4r+2}{r(r+1)(r+2)}=\frac{2r+1}{r(r+1)(r+2)},\)
as required.
(ii) Therefore
\(\sum_{r=1}^{n}\frac{2r+1}{r(r+1)(r+2)}\)
\(=\frac12\sum_{r=1}^{n}\left(\frac{(2r+1)(2r+3)}{(r+1)(r+2)}-\frac{(2r-1)(2r+1)}{r(r+1)}\right).\)
This is telescoping:
\(\frac12\Bigg[\left(\frac{3\cdot 5}{2\cdot 3}-\frac{1\cdot 3}{1\cdot 2}\right)+\left(\frac{5\cdot 7}{3\cdot 4}-\frac{3\cdot 5}{2\cdot 3}\right)+\cdots+\left(\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac{(2n-1)(2n+1)}{n(n+1)}\right)\Bigg].\)
All intermediate terms cancel, leaving
\(\frac12\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac{1\cdot 3}{1\cdot 2}\right\}.\)
Since \(\frac{1\cdot 3}{1\cdot 2}=\frac32\),
\(\sum_{r=1}^{n}\frac{2r+1}{r(r+1)(r+2)}=\frac12\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac32\right\}.\)
(iii) As \(n\to\infty\),
\(\frac{(2n+1)(2n+3)}{(n+1)(n+2)}\to 4.\)
So
\(\sum_{r=1}^{\infty}\frac{2r+1}{r(r+1)(r+2)}=\frac12\left(4-\frac32\right)=\frac54.\)
(i) Start with the right-hand side and simplify the numerator:
\(\frac{1}{2}\left\{\frac{(2r+1)(2r+3)}{(r+1)(r+2)}-\frac{(2r-1)(2r+1)}{r(r+1)}\right\}\)
\(=\frac{1}{2}\left\{\frac{r(2r+1)(2r+3)-(r+2)(2r-1)(2r+1)}{r(r+1)(r+2)}\right\}.\)
Now expand:
\(r(2r+1)(2r+3)=r(4r^2+8r+3)=4r^3+8r^2+3r,\)
\((r+2)(2r-1)(2r+1)=(r+2)(4r^2-1)=4r^3+8r^2-r-2.\)
Hence the numerator is
\(4r^3+8r^2+3r-(4r^3+8r^2-r-2)=4r+2=2(2r+1).\)
Therefore
\(\frac{1}{2}\left\{\frac{2(2r+1)}{r(r+1)(r+2)}\right\}=\frac{2r+1}{r(r+1)(r+2)}.\)
This verifies the identity.
(ii) Using part (i),
\(\sum_{r=1}^{n}\frac{2r+1}{r(r+1)(r+2)}=\frac12\sum_{r=1}^{n}\left(\frac{(2r+1)(2r+3)}{(r+1)(r+2)}-\frac{(2r-1)(2r+1)}{r(r+1)}\right).\)
Write out the first few terms:
\(\frac12\Bigg[\left(\frac{3\cdot 5}{2\cdot 3}-\frac{1\cdot 3}{1\cdot 2}\right)+\left(\frac{5\cdot 7}{3\cdot 4}-\frac{3\cdot 5}{2\cdot 3}\right)+\cdots+\left(\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac{(2n-1)(2n+1)}{n(n+1)}\right)\Bigg].\)
Everything in the middle cancels, leaving
\(\frac12\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac{1\cdot 3}{1\cdot 2}\right\}.\)
So
\(\sum_{r=1}^{n}\frac{2r+1}{r(r+1)(r+2)}=\frac12\left\{\frac{(2n+1)(2n+3)}{(n+1)(n+2)}-\frac32\right\}.\)
(iii) Now let \(n\to\infty\). Since
\(\frac{(2n+1)(2n+3)}{(n+1)(n+2)}\to \frac{4n^2}{n^2}=4,\)
the infinite sum is
\(\frac12\left(4-\frac32\right)=\frac54.\)
