Proof by induction - fractions

[mikky05v]

Homework Statement

I have been working on this proof for a few hours and I can not make it work out.

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

i need to get to $$1-\frac{1}{k+2}$$

I get as far as $$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
then I have tried $$1-\frac{(k+2)+1}{(k+1)(k+2)}$$ bu multiplying the left fraction by (k+2) which got me nowhere.

What am I doing wrong?

 

[vela]

Homework Statement

I have been working on this proof for a few hours and I can not make it work out.

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

i need to get to $$1-\frac{1}{k+2}$$

I get as far as $$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
then I have tried $$1-\frac{(k+2)+1}{(k+1)(k+2)}$$ bu multiplying the left fraction by (k+2) which got me nowhere.

What am I doing wrong?

You have a sign error. If you can't find it, try expanding
$$1-\frac{(k+2)+1}{(k+1)(k+2)}$$ and compare it to what you started with.