# Summation of a series giving logarithm

• Jul 26th 2010, 07:10 PM
Heirot
Summation of a series giving logarithm
I need to prove that $\sum_{i=1}^{\infty}\sum_{j=1}^{i-1}\frac{(-1)^i}{i j}=\frac{1}{2}\ln ^2 2$. The obvious way would be to compare with the series on the r.h.s. which is $\frac{1}{2}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\ frac{(-1)^{(i+j)}}{i j}$, but I just can't show that they're the same. Any suggestions?
• Jul 27th 2010, 06:01 AM
Renji Rodrigo
$\sum\limits^{\infty}_{k=1}\sum\limits^{k-1}_{j=1}\frac{(-1)^{k}}{(k)j}.$

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

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

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

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

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

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

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

$\int^{1}_{0}\frac{ln(x+1)}{x+1}=\frac{\ln^2 (2)}{2}.$
• Jul 27th 2010, 06:36 AM
Heirot
Thank you very much! Your method of dealing with sums has enlightened me.
• Jul 27th 2010, 07:15 AM
Renji Rodrigo
you're welcome! :)