<s>[INST] 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
 [/INST] 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 $n\neq -2$.
</s>