<s>[INST] Induction Proof for $\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)} = \frac{n}{(n+2)}$ I am asked: Prove, using induction, that for all $n \in \mathbb{N}$ the following equality holds.
  \begin{align*}
  \sum \limits_{i=1}^n \dfrac{2}{(i+1)(i+2)} = \dfrac{n}{(n+2)}
  \end{align*}
Here's my attempt:
For all $n\in \mathbb{N}$, let $P(n)$: $\sum \limits_{i=1}^n \frac{2}{(i+1)(i+1)}=\frac{n}{n+2}$.
Base Case: When $n=1$, $P(1)$: $\sum \limits_{i=1}^1 \frac{2}{(i+1)(i+2)}=\frac{1}{1+2}\implies \frac{2}{(1+1)(1+2)}=\frac{1}{3}\implies \frac{2}{6}=\frac{1}{3}\implies \frac{1}{3}=\frac{1}{3}$. Therefore, the base case is true.
Inductive Assumption: Let $n\in \mathbb{N}$ be generic and assume $P(n)$ is true, i.e. $\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)}=\frac{n}{n+2}$.
Induction Step: Prove $P(n+1)$ is true. $\sum \limits_{i=1}^{n+1} \frac{2}{(i+1)(i+2)}=\frac{2}{((n+1)+1)((n+2)+2)}+\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)}=\frac{2}{(n+2)(n+3)}+\frac{n}{n+2}=\frac{2}{(n+2)(n+3)}+\frac{n(n+3)}{(n+2)(n+3}=\frac{2+n(n+3)}{(n+2)(n+3)}=\frac{n^2+3n+2}{(n+2)(n+3)}=\frac{(n+1)(n+2)}{(n+2)(n+3)}=\frac{n+1}{n+3}=\frac{n+1}{(n+1)+2}$.
Therefore, $P(n+1)$ is true. By induction, $P(n)$ is true. $\blacksquare$
Did I miss anything? Thanks!
 [/INST] In the inductive step $$\sum \limits_{i=1}^{n+1} \frac{2}{(i+1)(i+2)}=\frac{2}{((n+1)+1)((n+2)+2)}+\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)}=$$
You meant $$\sum \limits_{i=1}^{n+1} \frac{2}{(i+1)(i+2)}=\frac{2}{((n+1)+1)((n+1)+2)}+\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)}=$$
The rest of your proof is flawless.
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