Proof by Induction - Math

Question

Prove that for every integer n>=2:

We have the summation of

$$ \frac{1}{i(i-1)}=1-\frac{1}{n} $$

I tried the algebra with this proof, but couldn't get it. I know that you split the i-1 and the i up but I am stuck from there.

Answer 1

The case $n=2$ is easy to verify. Now we assume we already know the case $n=k$, i.e., $$\sum_{i=2}^k \frac{1}{i(i-1)}=1-\frac{1}{k}$$ Now we look at case $n=k+1$. We expect to have $$\sum_{i=2}^{k+1} \frac{1}{i(i-1)}=1-\frac{1}{k+1}$$ We have $$ \sum_{i=2}^{k+1} \frac{1}{i(i-1)} =\sum_{i=2}^k \frac{1}{i(i-1)} + \frac{1}{(k+1)(k)}=1-\frac{1}{k}+ \frac{1}{(k+1)(k)}=1-\frac{1}{k+1}$$ Hence, the case $n=k+1$ is proved by using $n=k$. And this is how induction works

Answer 2

It is a telescopic sum, $\dfrac1{i(i-1)}=\dfrac1{i-1}-\dfrac1i$.