<s>[INST] Prove the following equality using mathematical induction: Prove the following equality using mathematical induction: For any integer $n \ge 1$
$$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$$
I understand for the base base I need to have $n=1$. If I substituted $n$ for 1 and $i$ for 1 I would get 
\begin{align}
\sum_{i=1}^{1} \frac{1}{1(1+1)} &= \frac{1}{1+1} \\
\sum_{i=1}^{1} \frac{1}{2} &= \frac{1}{2}
\end{align}
As for the inductive step, I inputted $n +1$ but it did not work for me because honestly I did not know what to do with the $i$ and what I should do next. 
How would I go about doing the next step and an explanation for $i$ would be helpful. 
 [/INST] Hint:
Write $$\frac{n}{n+1} = \frac{n+1-1}{n+1} = 1-\frac{1}{n+1}$$
And note that the sum on the left side is a telescop sum. Thus you do not even need induction to proof this. 

However, for the induction step, you could write
$$\begin{array}{rcl}
\sum_{i=1}^{n+1} \frac{1}{i(i+1)} &=& \frac{1}{(n+1)(n+2)}+\sum_{i=1}^{n} \frac{1}{i(i+1)} \\ 
&=& 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \\
&=& 1 - \frac{n+2}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)} \\
&=& 1 - \frac{n+2-1}{(n+1)(n+2)}\\
&=& 1 - \frac{1}{n+2}\\
\end{array}$$
</s>