Solve summation problem: $\sum_{k=1}^{n} \tfrac{2k+1}{k^{2}(k+1)^2 } $

Using partial fractions we look for $a,b,c,d$ such that 
\[\frac{2k+1}{k^2(k+1)^2}=\frac{a}{k}+\frac{b}{k^2}+\frac{c}{k+1}+\frac{d}{(k+1)^2}.          (1)\]
Multiply both sides of (1) by $k^2$ to get 
\[\frac{2k+1}{(k+1)^2}=ak+b+\frac{ck^2}{k+1}+\frac{d k^2}{(k+1)^2},\]
and substitute $k=0$ to get
\[1=b.\]
Multiply both sides of (1) by $(k+1)^2$ to get
\[\frac{2k+1}{k^2}=\frac{a}{k}(k+1)^2+\frac{b}{k^2}(k+1)^2+c(k+1)+d,\]
and substitute $k=-1$ to get
\[\frac{-1}{1}=d \Rightarrow d=-1.\]
Hence
\[\frac{2k+1}{k^2(k+1)^2}=\frac{a}{k}+\frac{1}{k^2}+\frac{c}{k+1}-\frac{1}{(k+1)^2}      \]
\[=\frac{ak(k+1)^2+(k+1)^2+ck^2(k+1)-k^2}{k^2(k+1)^2}\]
\[= \frac{ak(k^2+2k+1)+(k^2+2k+1)+ck^2(k+1)-k^2}{k^2(k+1)^2}\]
\[=\frac{(a+c)k^3+(2a+c)k^2+(a+2)k+1}{k^2(k+1)^2}=\frac{2k+1}{k^2(k+1)^2}.\]
Hence
\[a+c=0      2a+c=0,   \text{and}  a+2=2.\]
Thus $a=c=0$. Therefore
\[\frac{2k+1}{k^2(k+1)^2}=\frac{1}{k^2}-\frac{1}{(k+1)^2},\]
and hence
\[\sum_{k=1}^{\infty}\frac{2k+1}{k^2(k+1)^2}=\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{(k+1)^2}\]
\[=(1-\frac{1}{4})+(\frac{1}{4}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{16})+\dots=1.\]
So
\[\sum_{k=1}^{\infty}\frac{2k+1}{k^2(k+1)^2}=1.\]