Since $$ \sum^{n}_{r=1}\frac{3}{(3r-1)(3r+2)}=\frac{1}{2}-\frac{1}{3n+2} $$ find the sum of the series $$ \frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\cdots+\frac{1}{29\times32} $$ and find $$ \sum^{\infty}_{r=1}\frac{1}{(3r-1)(3r+2)} $$
I proved the first identity with partial fractions and the method of differences; and in order to work out the sum of the series I did:
$$ \sum^{10}_{r=1}=\frac{1}{2}-\frac{1}{32} = \frac{15}{32} $$
But the answer listed is $\frac{5}{32}$. In terms of the last problem, it seemed to me that as $r \rightarrow \infty$, the sum tended to $\frac{1}{2}$ or $\frac{1}{2}^{-}$. But the answer listed is $\frac{1}{6}$. Why is this and where am I going wrong with both?
$$\frac1{(3r-1)(3r+2)}=\frac13\frac{(3r+2)-(3r-1)}{(3r+2)(3r-1)}=\frac13\left(\frac1{3r-1}-\frac1{3r+2}\right)$$
You have missed the $\dfrac13$