Determine if $\frac{k-1}{k}+\frac{1}{k(k+1)}=\frac{k}{k+1}$ holds How to prove if the following equality holds? $$\frac{k-1}{k}+\frac{1}{k(k+1)}=\frac{k}{k+1}$$
Maybe finding a common denominator would work, but I have no idea how to do it in this example.
I see that it holds for $k=1$: $$0+\frac12=\frac12=\frac1{1+1}.$$
It also holds for $k=2$: $$\frac12+\frac16 = \frac36+\frac16 = \frac46 = \frac23 = \frac2{2+1}$$
It also works for $k=3$: $$\frac23+\frac1{12}=\frac8{12}+\frac1{12}=\frac9{12}=\frac34=\frac3{3+1}$$
But I am not sure how to proceed if I am not working with numbers but with expressions.
 A: Notice that $k(k+1)$ is a common multiple of $k$ and $k(k+1)$. 
Hence, for $k\neq 0$ and $k \neq -1$, we have
$$\frac{k-1}{k}+\frac{1}{k(k+1)}=\frac{k^2-1}{k(k+1)}+  \frac{1}{k(k+1)} = \frac{k^2}{k(k+1)} = \frac{k}{k+1}.$$
A: You're having trouble with finding the common denominator. Let's compare the two denominators.
$\frac{k-1}{k} + \frac{1}{k(k+1)}$ the difference between these two is one(on the right) is multiplied by $(k+1)$ and the other(the left) is not. To make the denominators common we multiply the numerator and the denominator of the left fraction by $\frac{k+1}{k+1}$ which if you think about it, is the same as multiplying it by one, which does not affect the value of the fraction. Doing this we get:
$\frac{k-1}{k} + \frac{1}{k(k+1)} = \frac{k^2-1}{k(k+1)} + \frac{1}{k(k+1)}$ $= \frac{k^2}{k(k+1)}$ (the $1$s cancel when we add them) $=\frac{k*k}{k(k+1)} = \frac{k}{k+1}$ (We cancel the $k$s on the top and bottom)
It is important that this equality holds almost everywhere. We cannot say it is valid at $k= 0,-1$ this is because these values of $k$ result in division by $0$ which we know is not allowed.
