Question

Prove that for any 0 < x < 1, the series (1−x)+(x²−x³)+... converges and show that it converges to 1/(1+x)

Answer 1

Prove that for any 0 < x < 1, the series (1−x)+(x²−x³)+... converges and show that it converges to 1/(1+x)

Solution:

$\displaystyle(1−x)+(x^2−x^3)+(x^4−x^5)+...=\sum_{\mathclap{n=0}} ^{\mathclap{\infty}}(1-x)x^{2n}$

$\displaystyle\sum_{\mathclap{n=0}} ^{\mathclap{\infty}}(1-x)x^{2n}=\displaystyle\sum_{\mathclap{n=0}} ^{\mathclap{\infty}}(1-x)({x^2})^n$

$\displaystyle\sum_{\mathclap{n=0}} ^{\mathclap{\infty}}(1-x)({x^2})^n$ is an infinite geometric series. It is an infinite series whose successive terms have a common ratio $x^2$ . Such a series converges if and only if the absolute value of the common ratio is less than one ($|x^2| < 1$ ).

Since:

$0 < x < 1$

then:

$0 < x^2 < 1$ and the convergence condition is satisfied.

The sum of the series can be computed from the finite sum formula:

$\displaystyle\sum_{\mathclap{n=0}} ^{\mathclap{\infty}}(1-x)({x^2})^n=\lim_{\mathclap{k\to\infty}}\displaystyle\sum_{\mathclap{n=0}} ^{\mathclap{k}}(1-x)({x^2})^n=$

$\displaystyle\lim_{\mathclap{k\to\infty}}\frac{(1-x)(1-(x^2)^{k+1})}{1-x^{2}}=\frac{1-x}{1-x^2}=$

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