Confusion about proving $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$ by induction I know what the answer to this question is, but I am not sure how the answer was reached and I would really like to understand it! I am omitting the base case because it is not relevant for my question.
Inductive hypothesis:
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$$ is true when $n = k$ and $k > 1$
Therefore: $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1}$$
Inductive step:
Prove that $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+1+1} = \frac{k+1}{k+2}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)}\right] + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+2}$$
What I am confused about is where the $\frac{1}{(k+1)(k+2)}$ comes from in the first line of the inductive step. Can someone please explain this in a little more detail? The source of the answer explains it as "break last term from sum", but I am unclear on what that means.

You want to show
$$\sum_{j=1}^n \frac{1}{j(j+1)} = \frac{n}{n+1}$$
The inductive step involves assuming it holds for $n=k$ and then showing that it also holds for $n=k+1$.  So you assume
$$\sum_{j=1}^k \frac{1}{j(j+1)} = \frac{k}{k+1}$$
and show
$$\sum_{j=1}^{k+1} \frac{1}{j(j+1)} = \frac{k+1}{(k+1)+1}.$$
The left side of the sum above can also be written like this:
$$\sum_{j=1}^{k+1} \frac{1}{j(j+1)} = \left[\sum_{j=1}^{k} \frac{1}{j(j+1)}\right] + \frac{1}{(k+1)(k+2)}.$$
This is breaking the last term from the sum. Now you can substitute in the inductive assumption for the sum in square brackets:
$$\sum_{j=1}^{k+1} \frac{1}{j(j+1)} = \left[\frac{k}{k+1}\right] + \frac{1}{(k+1)(k+2)}.$$
Now you need to show that
$$\frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2},$$
and you're done.
