<s>[INST] 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.
 [/INST] The inductive hypothesis is
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{k(k+1)}
=\frac{k}{k+1}.\tag1$$
You need to prove that (1) implies the statement got from (1) by replacing
$k$ by $k+1$. This is
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{(k+1)(k+2)}
=\frac{k+1}{k+2}\tag2$$
but instead of (2) you have
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{k(k+1)}
=\frac{k+1}{k+2}$$
which is wrong, and this is causing your confusion. The inductive
proof starts by recognising that
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{(k+1)(k+2)}
=\left[\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{k(k+1)}
\right]+\frac1{(k+1)(k+2)}.$$
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