# Some More Induction

1. Jun 26, 2008

### Moridin

1. The problem statement, all variables and given/known data

Show that

$$\forall n \in \matbb{N}:~~ \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$$

3. The attempt at a solution

(1) Show that it is true for n = 1:

$$\sum_{k=1}^{2} \frac{(-1)^{k+1}}{k} = \frac{(-1)^2}{1} +\frac{(-1)^3}{2} = 1 - 1/2 = 1/2$$

$$\sum_{k=n+1}^{2n} \frac{1}{k} = 1/2$$

(2) Show that if it is true for n = p, it is also true for n= p+1

Assume that

$$\sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} = \sum_{k=p+1}^{2p} \frac{1}{k}$$

Now,

$$\sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + .... - \frac{1}{2p}$$

$$\sum_{k=1}^{2(p+1)} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + .... - \frac{1}{2p} + \frac{1}{2p+1} - \frac{1}{2p+2} = \sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} + \frac{1}{2p+1} - \frac{1}{2p+2} = \sum_{k=p+1}^{2p} \frac{1}{k} + \frac{1}{2p+1} - \frac{1}{2p+2}$$

So, if it could be demonstrated that

$$\sum_{k=p+1}^{2p} \frac{1}{k} + \frac{1}{2p+1} - \frac{1}{2p+2} = \sum_{k=p+1}^{2(p+1)} \frac{1}{k}$$

then we are done with (2)? To a first approximation, this does not seem to be equal? I probably made a mistake somewhere.

2. Jun 26, 2008

### dirk_mec1

The sum should go to 2p+1 not 2(p+1) =2p+2, do you agree with me?

3. Jun 26, 2008

### Moridin

Yeah, that gets rid of the last negative term. Should the last sum go from k = p + 1 or k = p + 2?

4. Jun 26, 2008

### dirk_mec1

What do you think? How does one implement induction?

5. Jun 26, 2008

### Moridin

Well, it is true for k = p + 1, that's for sure. The rest I know how to do by just applying the axiom of induction.