$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$ by induction. I am wondering if I can get some help with this question.  I feel like this is false, as I have tried many ways even to get the base case working (for induction) and I can't seem to get it.  Can anyone confirm that this is false? If I am wrong, I would really appreciate a hint. 
$$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$$

Let $$F(n)=\sum_{i=1}^n\frac{1}{i(i+1)}$$
and $$G(n)=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$$
Your task is to prove that $F(n)=G(n)$ for all $n$.  To do this by induction, prove first that $F(1)=G(1)$.  Then, assume $F(n)=G(n)$.  Add $\frac{1}{(n+1)(n+2)}$ to both sides; this is because $F(n)+\frac{1}{(n+1)(n+2)}=F(n+1)$.  We now have $$F(n+1)=G(n)+\frac{1}{(n+1)(n+2)}$$
Now you need to do some algebra to prove that $G(n)+\frac{1}{(n+1)(n+2)}=G(n+1)$, at which point you're done.
