I have tried showing this by induction on $n$, but I got stuck. Here is what I had.
Base case: True for $n=1$. $$ \sum_{i=1}^1 \frac{1}{(1+i-1)(1+i)} = \frac{1}{1(1+1)} = \frac{1}{2 \cdot 1} $$
Assuming true for $n=k$, show for $n=k+1$. $$ \sum_{i=1}^{k+1} \frac{1}{((k+1)+i-1)((k+1)+i)} =\sum_{i=1}^{k+1} \frac{1}{(k+i)(k+i+1)} $$
But now I don't know how to separate the sum into "$\sum_{i=1}^k {(k+i-1)(k+i)}$ plus some extra" form.
If someone could point me in the right direction that would be great! Also, if anyone can suggest an alternate way to prove this, that would be helpful.
I figured it out with the help of lab bhattacharjee.
If I write the numerator as $(n+i) - (n+i-1)$, and if I define the sequence $\{a_i\} = \frac{1}{n+i}$, I can use the method of Telescoping Series so that the sum boils down to $$ \begin{align*} \sum_{i=1}^n {\frac{(n+i) - (n+i-1)}{(n+i-1)(n+i)}} &=\sum_{i=1}^n \left( {\frac{n+i}{(n+i-1)(n+i)} - \frac{n+i-1}{(n+i-1)(n+i)}}\right)\\ &=\sum_{i=1}^n{\left( \frac{1}{n+i-1} - \frac{1}{n+i} \right)}\\ &= \sum_{i=1}^n { ( a_{i-1} - a_i ) }\\ &= -\sum_{i=1}^n {(a_i - a_{i-1})}\\ &= -(a_n - a_0)\\ &= \frac{1}{n+0} - \frac{1}{n+n}\\ &= \frac{1}{2n}. \end{align*} $$