Prove $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$ using induction.

For $n\geq 2$, let $S(n)$ denote the statement $$ S(n) : \sum_{i=2}^n \frac{1}{(i-1)i}=\frac{n-1}{n}. $$ Base step ($n=2$): $S(2)$ says that $\frac{1}{(2-1)2}=\frac{1}{2}=\frac{2-1}{2}$, and this is true.

Inductive step: Fix some $k\geq 2$ and assume that $S(k)$ is true where $$ S(k) : \sum_{i=2}^k \frac{1}{(i-1)i}=\frac{k-1}{k}. $$ To be shown is that $S(k+1)$ follows where $$ S(k+1) : \sum_{i=2}^{k+1} \frac{1}{(i-1)i}=\frac{k}{k+1}. $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} \sum_{i=2}^{k+1} \frac{1}{(i-1)i} &= \frac{1}{k(k+1)}+\sum_{i=2}^{k}\frac{1}{(i-1)i}\tag{by defn. of $\Sigma$}\\[1em] &= \frac{1}{k(k+1)}+\frac{k-1}{k}\tag{by ind. hyp.}\\[1em] &= \frac{1+(k-1)(k+1)}{k(k+1)}\tag{common denom}\\[1em] &= \frac{k^2}{k(k+1)}\tag{simplify}\\[1em] &= \frac{k}{k+1}\tag{desired expression}, \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.

Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 2$. $\blacksquare$

Assume $\sum\limits_{i=2}^n\frac{1}{(i-1)i}=\frac{n-1}{n}$. Then $$\sum_{i=2}^{n+1}\frac{1}{(i-1)i}=\sum_{i=2}^n\frac{1}{(i-1)i}+\frac{1}{n(n+1)}\stackrel{\text{ind. hyp.}}=\frac{n-1}{n}+\frac{1}{n(n+1)}=$$$$=\frac{(n-1)(n+1)+1}{n(n+1)}=\frac{n^2-1+1}{n(n+1)}=\frac{n^2}{n(n+1)}=\frac{n}{n+1}=\frac{(n+1)-1}{n+1}$$

Your problem was that you added $\frac{1}{n^2}$ instead of $\frac{1}{n(n+1)}$. You want what you add to be the $n+1$'th member of the sequence $a_n=\frac{1}{(n-1)n}$ so that you find $\sum\limits_{i=2}^{n+1}a_i$ (which is $\sum\limits_{i=2}^n a_i+a_{n+1}$) given $\sum\limits_{i=2}^n a_i$.

You should also denote the sum by $\sum\limits_{i=2}^n \frac{1}{(i-1)i}$, not $\sum\limits_{i=2}^n \frac{1}{(n-1)n}$.

In addition to that, you later claimed that $\frac{n}{n+1}=n+1$, which is false.