0.999.../The geometric series formula

$\sum _{n=0}^{\infty }x^{n}=1+x+x^{2}+x^{3}+...$ is known as a geometric series.

If, $|x|<1$, then this series converges to:

$\sum _{n=0}^{\infty }x^{n}={\frac {1}{1-x}}$

Proof: Define the partial sum $S_{n}$:

$S_{n}=\sum _{n=0}^{n}=1+x+x^{2}+x^{3}+...x^{n-1}$

$xS_{n}=\sum _{n=0}^{n}=..+x+x^{2}+x^{3}+...x^{n-1}+x^{n}$

Note that both partial sums have n terms. When they are subtracted only the first term in and the last term in $xS_{n}$ will remain:

$S_{n}-xS_{n}=1-x^{n}$, which can be easily solved for $S_{n}$

Note that we have failed to establish when the infinite series converges. This requires an understanding of the what happens when we take the limit of the partial sum as n goes to infinity. This is left to the reader as problem 1.

Sample problems:

1. Prove that sequence of partial sums, $\{S_{n}\}_{n=1}^{\infty }=\{S_{1},S_{2},S_{3},...\}$, converges if $|x|<1$.
2. This can be used to convert other repeating decimals into fractions. For pedagogical purposes, it is better to begin with a known fraction. From such a list, we consider: 0.181818... . Convert this into a rational fraction.

Footnotes

• See also Calculus/Sequences and Series/Exercises