Simplifying fractions - Ending up with wrong sign I've been trying to simplify this
$$
1-\frac{1}{n+2}+\frac{1}{(n+2) (n+3)}
$$
to get it to that
$$
1-\frac{(n+3)-1}{(n+2)(n+3)}
$$
but I always end up with this
$$
1-\frac{(n+3)+1}{(n+2)(n+3)}
$$
Any ideas of where I'm going wrong?
Wolfram Alpha gets it to correct form but it doesn't show me the steps (even in pro version)
Thanks

Just in case you want to see a full simplification: 
\begin{align*}
1-\frac{1}{n+2}+\frac{1}{(n+2)(n+3)} &= \frac{(n+2)(n+3)}{(n+2)(n+3)} - \frac{(n+3)}{(n+2)(n+3)}+\frac{1}{(n+2)(n+3)} \\
&= \frac{(n+2)(n+3)-(n+3)+1}{(n+2)(n+3)}\\
&= \frac{(n^2+5n+6) -n-2 }{(n+2)(n+3)} \\
&= \frac{n^2+4n+4}{(n+2)(n+3)} \\
&= \frac{(n+2)^2}{(n+2)(n+3)} \\
&= \frac{n+2}{n+3} \\
\end{align*}
Provided $n\neq -2$.
