jmc

First, we split $\frac{2n-1}{n(n+1)(n+2)}$ into partial fractions by writing $$\frac{2n-1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}.$$Then $2n-1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1).$

Setting $n=0,$ we get $-1=2A,$ so $A=-\frac{1}{2}.$

Setting $n=-1,$ we get $-3=-B,$ so $B=3.$

Setting $n=-2,$ we get $2C=-5,$ so $C=-\frac{5}{2}.$ Thus, $$\frac{2n-1}{n(n+1)(n+2)}=-\frac{1/2}{n}+\frac{3}{n+1}-\frac{5/2}{n+2}.$$Therefore, $$\begin{array}{rl}\sum _{n=1}^{\infty }\frac{2n-1}{n(n+1)(n+2)}& =\left(-\frac{1/2}{1}+\frac{3}{2}-\frac{5/2}{3}\right)+\left(-\frac{1/2}{2}+\frac{3}{3}-\frac{5/2}{4}\right)\\ & +\left(-\frac{1/2}{3}+\frac{3}{4}-\frac{5/2}{5}\right)+\left(-\frac{1/2}{4}+\frac{3}{5}-\frac{5/2}{6}\right)+\cdots \\ & =-\frac{1}{2}+\frac{5/2}{2}=\boxed{\frac{3}{4}}.\end{array}$$