$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$ 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. 
$$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$$

For $n=1$, we have
$$\frac{1}{2} = \sum_{i=1}^1 \frac{1}{i(i+1)} = \sum_{i=1}^n \frac{1}{i(i+1)} \ne \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)} = \frac{3}{4} - \frac{2+3}{2(1+1)(1+2)} = \frac{1}{3}.$$
So, yes, I'd say you're right.
