## Saturday, April 16, 2011

### Cauchy Product (1 − 1 + 1 − 1 + ... )

Perquisites :

given

$1-1+1-1+1-1\cdot \cdot \cdot =\sum _{n=0}^{\infty }a_{n}=\sum _{n=0}^{\infty }b_{n}=\sum _{n=0}^{\infty }(-1)^{n}$

Cauchy product
$\begin{array}{rcl} c_n & = &\displaystyle \sum_{k=0}^n a_k b_{n-k}=\sum_{k=0}^n (-1)^k (-1)^{n-k} \\[1em] & = &\displaystyle \sum_{k=0}^n (-1)^n = (-1)^n(n+1). \end{array}$
then, the product is
$\sum_{n=0}^\infty(-1)^n(n+1) = 1-2+3-4+\cdots.$
therefore,
$\left ( 1-1+1-1+1-1\cdot \cdot \cdot \right )\left ( 1-1+1-1+1-1\cdot \cdot \cdot \right )=\left ( 1-2+3-4+5-6\cdot \cdot \cdot \right )$
(1)
based  on geometric series (GS.ID.1)
$\sum_{k=0}^{n} a r^k = \frac{a}{1-r}.$
(2)
a=1 , r=-1 in the following equation
$1-1+1-1+1-1\cdot \cdot \cdot =\sum _{n=0}^{\infty }(-1)^{n}$
(3)
from (2) and (3)
$1-1+1-1+1-1\cdot \cdot \cdot =\frac{1}{2}$
(4)

from (1) and (4)
$1-2+3-4+5-6\cdot \cdot \cdot =\frac{1}{4}$