Question

Show that
$\frac{1}{a}-\frac{1}{ab}+\frac{1}{abc}-\frac{1}{abcd}+\cdots =\frac{1}{a+}\frac{a}{b-1+}\frac{b}{c-1+}\frac{c}{d-1+}\cdots$.

Answer 1

$\Rightarrow$ Let $\frac{1}{a}-\frac{1}{ab}=\frac{1}{a+\beta };$ thus $(b-1)(a+\beta )=ab$

and $\beta =\frac{a}{b-1}$

$\therefore \frac{1}{a}-\frac{1}{ab}=\frac{1}{a+}\frac{a}{b-1}$

Similarly, $\frac{1}{a}-\frac{1}{ab}+\frac{1}{abc}=\frac{1}{a}-\frac{1}{a}(\frac{1}{b}-\frac{1}{bc})$$=\frac{1}{a}-\frac{1}{a(b+\gamma )}$

$=\frac{1}{a+}\frac{1}{b-1+\gamma }=\frac{1}{a}\frac{a}{b-1+}\frac{b}{c-1+}.....$

Hence, $\frac{1}{a}-\frac{1}{ab}+\frac{1}{abc}-\frac{1}{abcd}...$

$=\frac{1}{a}\frac{a}{b-1+}\frac{b}{c-1+}\frac{c}{d-1+}.....$