# Inductive proof of summation formula

1. Oct 21, 2009

### livcon

1. The problem statement, all variables and given/known data
Prove by induction the following summation formula:
$$\frac{1}{1\1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
$$n \geq 1$$
2. Relevant equations
-
3. The attempt at a solution
Inductive step:
1. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{(n+1)+1}$$

2. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$

To prove this (1) should equal (2) (right?), but I don't see how to manipulate (2) to make it equal to (1)

2. Oct 21, 2009

### whs

Common denominator!

3. Oct 21, 2009

### Hurkyl

Staff Emeritus
You don't need to. All you need to do is to show these two sums of fractions are equal.

4. Oct 21, 2009

### livcon

That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks

5. Oct 21, 2009

### Hurkyl

Staff Emeritus
As a warmup exercise, prove the following:

$$\frac{1}{3} + \frac{1}{6} = \frac{3}{2} - 1$$

6. Oct 21, 2009

### emyt

see how 1 - 1/n+1 looks first

7. Oct 22, 2009

### livcon

I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.

8. Oct 22, 2009

### Hurkyl

Staff Emeritus
Isn't proving
$$1 - \frac{1}{(n+1)+1} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$​
exactly the same kind of problem, though? Why can't you do the same thing you did for the warmup problem?

9. Oct 22, 2009

### livcon

I guess it's because in the warmup exercise I could easily add and subtract the fractions because it was all constants, but in the one involving the variable n I got stuck. So I would probably have an epiphany if someone would just write it out. This is just a matter of me not seeing the obvious steps in this particular problem.

10. Oct 22, 2009

### HallsofIvy

Staff Emeritus
What do you get when you change
$$1- \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}$$
to fractions having the same denominator?

That's what everyone has been telling you do to. Have you done it?