# Sum of a series resulting in a logarithm

1. Jul 27, 2010

### Heirot

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

Prove: $$\sum_{i=1}^{\infty}\sum_{j=1}^{i-1} \frac{(-1)^i}{i j}=\frac{1}{2}\ln^2 2$$

2. Relevant equations

$$\ln 2 = \sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i}$$

3. The attempt at a solution

I'm trying to manipulate the l.h.s. of the problem to transform it into $$\frac{1}{2}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \frac{(-1)^{i+j}}{i j}$$ but I just can't seem to get it.

Any suggestions will be appreciated.

2. Jul 28, 2010

### sshzp4

I could give you the solution or I could give you a hint. Let me give you the hint first. The hint is:
log (1+x) = x -x^2/2 + x^3/3 - x^4/4 + x^5/5.. (ad nauseum to infinity)= $\int \frac{1}{1+x}dx$.

Substituting $x=1$ gives you your Eq. 2. You seem to have been thinking only symbolically so far, and that's why you are at wit's end. Now try thinking about what summations over variables really mean.

In your question, you have two summations, or alternatively two integrals; one in a dummy variable (let's call it $r$), the other in $x$. Remember that summations are discrete representations of integrals (with dx=a unit step).

$int \frac{1}{1+x}\left[ \int^{x} \frac{1}{1-r}dr \right] dx$

There's a mistake in the above equation for you to take note of but you should have sufficient direction to move forward. So you might ask how can we take the $\frac{(-1)^i}{i}$ out of the term... that's only because $\sum_j$ does not affect $i$, so it's effectively a constant under that summation. Also look up the term 'dilogarithms'.

I will post the solution after I get to work, which should be around two hours from now, unless there's dead deers or rednecks blocking the traffic.

Last edited: Jul 28, 2010
3. Jul 28, 2010

### vela

Staff Emeritus
You should review the forum rules. Posting solutions isn't allowed here.

4. Jul 28, 2010

### sshzp4

Thanks vela. I wasn't aware of that.

Okay Heirot. You are on your own there now. The final hints:
1. You need to use the transform $j=r+1$. (You might want to check if the i-1 on top of your second summation sign in your Eq. 1, it is likely that that could be a i+1 instead). By no means is this the only transform that you use.

2. You end up with an integral expression that is expressed in terms of itself. I don't know the technical descriptor for this. eg. $I_n= -\frac{n!!}{(n-1)!!}\sin^n(x) \cos(x) -nI_n$ where $n$ is a constant and $I_n$ is the integral equation. Then you can solve the whole thing and substitute $x=1$ to get the final solution and then go and watch a movie.

(I am too lazy/find it intellectually abhorrent to read anything that can be described as 'rules of behavior', my personal failing obviously.)

5. Jul 28, 2010

### Heirot

Oh, I see, this can't be proven the way I was doing it. Thanks for your help!