Question

If x is a proper fraction, show that $\frac{x}{1-x^2}-\frac{x^3}{1-x^6}+\frac{x^5}{1-x^{10}}-.......=\frac{x}{1+x^2}+\frac{x^3}{1+x^6}+\frac{x^5}{1+x^{10}}+.....$

Answer 1

$\frac{x}{1+x^2}=x{(1-x^2)}^{-1}$

$=x(1+x^2+x^4+....)$

$=x+x^3+x^5+x^7+.......$

$\therefore \frac{x^3}{1-x^6}=x^3{(1-x^6)}^{-1}$

$=x^3+x^9+x^{15}+x^{21}+....$

Similarly;

$\frac{x}{1-x^2}-\frac{x^3}{1-x^6}+\frac{x^5}{1-x^{10}}-\frac{x^7}{1-x^{14}}+....$

$=x+x^3+x^5+x^7+.....$

$-x^3-x^9-x^{15}-x^{21}-....$

$+x^5+x^{15}+x^{25}+x^{35}+....$

$-x^7-x^{21}-x^{35}-x^{19}-....$

....................

By adding vertical columns;-

$=x(1-x^2+x^4-x^6....)+x^3(1-x^6+x^{12}-x^{18}+....)+x^5(1-x^{10}+x^{20}-x^{30}+....)$

$=x{(1+x^2)}^{-1}+x^3{(1+x^6)}^{-1}+x^5{(1+x^{10})}^{-1}+x^7{(1+x^{14})}^{-1}+....$

$=\frac{x}{1+x^2}+\frac{x^3}{1+x^6}+\frac{x^5}{1+x^{10}}+\frac{x^7}{1+x^{14}}+......$