$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$ 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. 
$$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$$
 A: Indeed, you're suspicion is correct. The given equality is false.
Just testing out $n = 1$ gives us a sum of $\dfrac 12 \neq \dfrac 34 - \dfrac 5{12}$.
What is true is that: $$\sum_{i = 1}^n \dfrac 1{i(i+1)} = \dfrac{n}{n+1} = 1 -\dfrac 1{n+1}\tag{$\star$}$$
Suggestion: Use induction on $n$ to verify $(\star)$
A: You’re quite right: clearly $$\sum_{i=1}^1\frac1{i(i+1)}=\frac12\ne\frac13=\frac34-\frac5{12}\;,$$ so you cannot even establish the base case.
In fact
$$\sum_{i=1}^n\frac1{i(i+1)}=\sum_{i=1}^n\left(\frac1i-\frac1{i+1}\right)=1-\frac1{n+1}=\frac{n}{n+1}\;,$$
while 
$$\begin{align*}
\frac34-\frac{2n+3}{2(n+1)(n+2)}&=\frac{3(n+1)(n+2)-2(2n+3)}{4(n+1)(n+2)}\\\\
&=\frac{n(3n+5)}{4(n+1)(n+2)}\\\\
&=\left(\frac{n}{n+1}\right)\left(\frac{3n+5}{4(n+2)}\right)\;,
\end{align*}$$
so the two are equal if and only if $3n+5=4n+8$, or $n=-3$. Thus, the two are never equal for any positive integer $n$.
A: note
$$\dfrac{1}{i(i+1)}=\dfrac{1}{i}-\dfrac{1}{i+1}$$
so
$$\sum_{i=1}^{n}\dfrac{1}{i(i+1)}=\dfrac{n}{n+1}$$
and so you can use induction easy to solve it
A: 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.
A: 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.
