First show it's true for [tex]n=1[/tex]. On the left,
[tex]\dfrac1{1\cdot2}=\dfrac12[/tex]
On the right,
[tex]\dfrac1{1+1}=\dfrac12[/tex]
so the base case [tex]n=1[/tex] is true.
Now assume equality holds for [tex]n=k[/tex], that
[tex]\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots\dfrac1{k(k+1)}=\dfrac k{k+1}[/tex]
We use this assumption to show it also holds for [tex]n=k+1[/tex]. By hypothesis,
[tex]\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots+\dfrac1{k(k+1)}+\dfrac1{(k+1)(k+2)}=\dfrac k{k+1}+\dfrac1{(k+1)(k+2)}[/tex]
(the first [tex]k[/tex] terms condense to [tex]\dfrac1{k(k+1)}[/tex])
Combining the fractions gives
[tex]\dfrac{k(k+2)}{(k+1)(k+2)}+\dfrac1{(k+1)(k+2)}=\dfrac{k^2+2k+1}{(k+1)(k+2)}=\dfrac{(k+1)^2}{(k+1)(k+2)}=\dfrac{k+1}{k+2}[/tex]
which is what we had to establish, thus proving (by induction) equality for all [tex]n[/tex].
