<s>[INST] Prove the following using induction $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
I'm new to induction, but this is what I cam up with so far.
$$1 - \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{k+2+k}{k(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{2(k+1)}{k(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
$$1 - \frac{2}{k(k+2)} = 1 - \frac{1}{k+2}$$
Where did I go wrong, or where do I go from here?
EDIT:
After some advice from Patrick (you're awesome!) I have...
$$1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$$
    $$1 - \frac{k + 2 + 1}{(k+1)(k+2)}$$
    $$1 - \frac{k + 3}{(k+1)(k+2)}$$
    $$\frac{(k+1)(k+2) - (k+3)}{(k+1)(k+2)}$$
    $$\frac{k^2 + 2k + k + 2 - k - 3)}{(k+1)(k+2)}$$
    $$\frac{k^2 + 2k - 1}{(k+1)(k+2)}$$
    $$\frac{k^2 + 2k - 1}{k^2 + 3k + 2}$$
Did I goof up somewhere again?
Thanks.
 [/INST] For every $n\in\mathbb{N}$, let $\mathcal{P}(n)$ the statement
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}.
$$


*

*The statement $\mathcal{P}(1)$ is $\frac{1}{1\times 2} = 1 - \frac{1}{2}$, which is clearly true.

*We prove that $\mathcal{P}(n) \implies \mathcal{P}(n+1)$  for all $n\geq 1$. We use a direct proof. Suppose
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}.
$$
Then
$$
\begin{align*}
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} 
 &= 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \\
 &= 1 - \frac{(n+2) - 1}{(n+1)(n+2)} \\
 &= 1 - \frac{1}{n+2}.
\end{align*}
$$
This proves that $\mathcal{P}(n) \implies \mathcal{P}(n+1)$. 


It follows by induction that
$$
\frac{1}{1\times 2} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}
$$
for all $n\geq 1$.
</s>