First show it's true for
. On the left,
![\frac1{1\cdot2}=\frac12](https://img.qammunity.org/2020/formulas/mathematics/college/i5i8t3wae0zqfk18yjln841xdr7pus06mq.png)
On the right,
![\frac1{1+1}=\frac12](https://img.qammunity.org/2020/formulas/mathematics/college/jnx676l1yfqv9qeag8kwysowi9j2zvxmha.png)
so the base case
is true.
Now assume equality holds for
, that
![\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots\frac1{k(k+1)}=\frac k{k+1}](https://img.qammunity.org/2020/formulas/mathematics/college/ebxvqxkhvxnjbn1vu1ap04fgao84t7zoo6.png)
We use this assumption to show it also holds for
. By hypothesis,
![\frac1{1\cdot2}+\frac1{2\cdot3}+\cdots+\frac1{k(k+1)}+\frac1{(k+1)(k+2)}=\frac k{k+1}+\frac1{(k+1)(k+2)}](https://img.qammunity.org/2020/formulas/mathematics/college/uufzaburec2af8fx81qz7iqc6hx6lvtcv0.png)
(the first
terms condense to
)
Combining the fractions gives
![(k(k+2))/((k+1)(k+2))+\frac1{(k+1)(k+2)}=(k^2+2k+1)/((k+1)(k+2))=((k+1)^2)/((k+1)(k+2))=(k+1)/(k+2)](https://img.qammunity.org/2020/formulas/mathematics/college/8hefb0ok19lj9bqsyqvb3nx5z9bddd6ijj.png)
which is what we had to establish, thus proving (by induction) equality for all
.