The coefficients a_n of sum from n=1 to infinity a_nx^n
The coefficients $a_n$ of $\sum_{n=1}^{\infty}a_nx^n$ are the successive decimal digits of $\dfrac47$. Determine the radius of convergence and the sum.
Explanation of Method
Since $4/7$ has a repeating decimal expansion, the coefficients cycle with a fixed period; grouping the series into blocks of that period turns it into a single polynomial times a geometric series in $x^{\text{period}}$.
- Convert $4/7$ to its repeating decimal and read off the digit cycle.
- Group the series into blocks of six and factor out the geometric part.
- Determine the radius of convergence and verify at $x=1/10$.
Step 1 · Decimal expansion
$$ \frac47 = 0.\overline{571428}, $$ a repeating decimal with period $6$: the digit cycle is $5,7,1,4,2,8$, repeating indefinitely ($a_{n+6}=a_n$).
Step 2 · Group into blocks of six
$$ \sum_{n=1}^{\infty}a_nx^n = \bigl(5x+7x^2+x^3+4x^4+2x^5+8x^6\bigr)\bigl(1+x^6+x^{12}+\cdots\bigr) = \frac{5x+7x^2+x^3+4x^4+2x^5+8x^6}{1-x^6}. $$
Step 3 · Radius of convergence and check
Since every $a_n$ is a bounded digit ($0$ through $9$), and the geometric factor $1/(1-x^6)$ converges precisely for $|x|<1$: $$ R=1. $$ At $x=\frac1{10}$: $$ S\!\left(\frac1{10}\right) = \frac{0.571428}{1-10^{-6}} = \frac{571428}{999999} = \frac47, $$ confirming the construction.
Summary Table
| Property | Value |
|---|---|
| Digit cycle | $5,7,1,4,2,8$ (period $6$) |
| Radius of convergence | $R=1$ |
| Sum | $\dfrac{5x+7x^2+x^3+4x^4+2x^5+8x^6}{1-x^6}$ |
Additional Notes
- This construction works for the repeating decimal of any rational number: a period-$p$ repeating decimal always produces coefficients satisfying $a_{n+p}=a_n$, and the same block-grouping trick applies, with $p$ replaced by the period of that fraction.
- The value $x=1/10$ recovers the original fraction exactly, since a decimal expansion is itself just this power series evaluated at $x=1/10$ — the construction is really "decimal expansion made into a formal power series."
- The radius of convergence, $1$, is far larger than the "natural" evaluation point $x=1/10$; the boundedness of digit coefficients (always between $0$ and $9$) is what guarantees convergence all the way out to $|x|=1$, regardless of which fraction's digits are used.