How do we get the k/(k+1) on the left hand side of the equation via the induction hypothesis? Do I have to change the right hand side at all or is it just (k+1)/(k+2)?

Hi baker11108,

**P(k)**
\(\displaystyle \frac{1}{2}+\frac{1}{6}+......+\frac{1}{k(k+1)}=\frac{k}{k+1}\)

**P(k+1)**
\(\displaystyle \frac{1}{2}+\frac{1}{6}+....+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}\)

You want to try to prove that the formula being true for n=k

__causes__ the formula to be true for n=k+1

Hence we write P(k+1) in terms of P(k) to see if there is a cause.

Therefore if the sum of the first k terms really is \(\displaystyle \frac{k}{k+1}\)

then the sum of k+1 terms is

\(\displaystyle \frac{1}{2}+\frac{1}{6}+......+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}\)

will be \(\displaystyle \frac{k}{k+1}+\frac{1}{(k+1)(k+2)}\)

which is

\(\displaystyle \left(\frac{k+2}{k+2}\right)\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}\)

\(\displaystyle =\frac{k^2+2k+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}\)

Since this is what you get when you replace k with k+1 in the RHS

it means that your hypothesis being true for some n=k

causes the hypothesis to be true for the next n=k+1.

Hence "true for n=1 causes it to be true for n=2 causes it to be true for n=3 causes......."