Thread: Summation of a series giving logarithm

1. Summation of a series giving logarithm

I need to prove that $\displaystyle \sum_{i=1}^{\infty}\sum_{j=1}^{i-1}\frac{(-1)^i}{i j}=\frac{1}{2}\ln ^2 2$. The obvious way would be to compare with the series on the r.h.s. which is $\displaystyle \frac{1}{2}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\ frac{(-1)^{(i+j)}}{i j}$, but I just can't show that they're the same. Any suggestions?

2. $\displaystyle \sum\limits^{\infty}_{k=1}\sum\limits^{k-1}_{j=1}\frac{(-1)^{k}}{(k)j}.$

We have
$\displaystyle \sum\limits^{\infty}_{k=1}\sum\limits^{k-1}_{j=1}\frac{(-1)^{k}}{(k)j}=\sum\limits^{\infty}_{k=0}\sum\limit s^{k}_{j=1}\frac{(-1)^{k+1}}{(k+1)j}.$

when $\displaystyle k=0$in the second series we have $\displaystyle \sum\limits^{k}_{j=1}\frac{(-1)^{k+1}}{(k+1)j}=0$ because the sum is empty, so we can write

$\displaystyle \sum\limits^{\infty}_{k=1}\sum\limits^{k}_{j=1}\fr ac{(-1)^{k+1}}{(k+1)j}$
reverting the order of summation

$\displaystyle \sum\limits^{\infty}_{k=1}\sum\limits^{k}_{j=1}\fr ac{(-1)^{k+1}}{(k+1)j}= \sum\limits^{\infty}_{j=1}\sum\limits^{\infty}_{k= j}\frac{(-1)^{k+1}}{(k+1)j}=$

$\displaystyle =\sum\limits^{\infty}_{j=1}\sum\limits^{\infty}_{k =0}\frac{(-1)^{k+1+j}}{(k+1+j)j}=$

$\displaystyle = \sum\limits^{\infty}_{j=1}\frac{(-1)^{j+1}}{j}\sum\limits^{\infty}_{k=0}(-1)^k \int^{1}_{0}x^{k+j}dx=$

$\displaystyle =\int^{1}_{0} \sum\limits^{\infty}_{j=1}\frac{(-1)^{j+1}x^j}{j}\sum\limits^{\infty}_{k=0} (-x)^{k}dx=$

$\displaystyle \int^{1}_{0}\frac{ln(x+1)}{x+1}=\frac{\ln^2 (2)}{2}.$

3. Thank you very much! Your method of dealing with sums has enlightened me.

4. you're welcome!