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.
 A: 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.
A: The base case $n=1$ is true.  Assume that for some $n> 1$ 
$$\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}$$
Then, we have
$$\begin{align}
\sum_{k=1}^{n+1}\frac{1}{k(k+1)}&=\sum_{k=1}^n\frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)}\\\\
&=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}\\\\
&=\frac{n(n+2)+1}{(n+1)(n+2)}\\\\
&=\frac{(n+1)^2}{(n+1)(n+2)}\\\\
&=\frac{n+1}{n+2}
\end{align}$$
which completes the proof by induction.
A: 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
$$\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{(k+1)(k+2)}
=\left[\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac{1}{k(k+1)}
\right]+\frac1{(k+1)(k+2)}.$$
A: "What I am confused about is where the 1/(k+1)*(k+2) comes from in the first line of the inductive step."
It comes because you are trying to evaluate for $n = k+1$
You want to prove the statement  $\frac 1{1*2} + .....+ \frac 1{(n-1)(n)} + \frac 1{n*(n+1)} = \frac n{n+1}$.
For $n = k$  if you replace $n$ with $k$ you are assuming $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)} = \frac k{k+1}$.
If you replace $n$ with $k + 1$ you get the statement:
$\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)}+ \frac 1{(k+1)(k+1)} = \frac {k+1}{k+2}$.
And that is the statement you wish to prove.  
We are assuming  $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)} = \frac k{k+1}$
So $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)}+ \frac 1{(k+1)(k+1)} = \frac k{k+1} + \frac 1{(k+1)(k+1)}$
You just need to prove that $\frac k{k+1} + \frac 1{(k+1)(k+1)}= \frac {k+1}{k+2}$.
.....
Anyway, you certainly did not transcribe to proof correctly.
