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 $k=1$

$0+\frac{1}{2}=\frac{1}{2}=\frac{1}{1+1}.$

It also holds for $k=2$

$\frac{1}{2}+\frac{1}{6}=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}=\frac{2}{2+1}$

It also works for $k=3$

$\frac{2}{3}+\frac{1}{12}=\frac{8}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}=\frac{3}{3+1}$

But I am not sure how to proceed if I am not working with numbers but with expressions.

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 $k=1$

$0+\frac{1}{2}=\frac{1}{2}=\frac{1}{1+1}.$

It also holds for $k=2$

$\frac{1}{2}+\frac{1}{6}=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}=\frac{2}{2+1}$

It also works for $k=3$

$\frac{2}{3}+\frac{1}{12}=\frac{8}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}=\frac{3}{3+1}$

But I am not sure how to proceed if I am not working with numbers but with expressions.