jmc

First, we decompose $\frac{2n+1}{n(n+1)(n+2)}$ into partial fractions. Let $$\frac{2n+1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}.$$Then $$2n+1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1).$$Setting $n=0,$ we get $2A=1,$ so $A=\frac{1}{2}.$

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

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