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What are some infinite series expansion for $3/7$ (and in general, for fractions with digits in base 10)? I can't think of something useful at all. Please generalize some useful series expression for all those kinds of fraction.

Can someone help?

  • 8
    I don't understand the question. Long division already provides such an expansion.2012-05-30
  • 5
    $0.428571428571428571428571\dots$. But you know this. What is the question?2012-05-30
  • 1
    What exactly is the point of this question? It would be useful if you gave some context, because as Marvis shows, there is an obvious answer that doesn't really tell us anything.2012-05-31
  • 2
    $\frac{3}{7}= \frac{3}{7} + 0 +0+0+ ..$2012-05-31

6 Answers 6

20

One might also try the mundane

$$\frac{3}{7} = \frac{1}{{2 + \frac{1}{3}}} = \frac{1}{2}\frac{1}{{1 + \frac{1}{6}}} = \frac{1}{2}\left( {1 - \frac{1}{6} + \frac{1}{{{6^2}}} - \frac{1}{{{6^3}}} + - \cdots } \right)$$

  • 5
    Which now that I look again is rather diabolic...2012-05-30
14

Consider any convergent series. Lets say that $$a_1 + a_2 + a_3 + \cdots = a$$ where $a_n, a \in \mathbb{R}$ and $a \neq 0$ (Thanks, @anon).

Define $b_n = \dfrac37\dfrac{a_n}{a}$, then the series $b_1 + b_2 + b_3 + \cdots$ converges to $\dfrac37$.

  • 3
    I think Marvis wins the cake...2012-05-31
10

Find the smallest $n$ such that $7|(10^n-1)$. In particular say $10^n-1=7m$. Then write

$$\begin{array}{c l} \frac{3}{7} & =\frac{3m}{7m} \\[2pt] & =\frac{3m}{10^n-1} \\[2pt] & = 3m\frac{1}{10^n}\frac{1}{1-1/10^n} \\[2pt] & =\frac{3m}{10^n}+\frac{3m}{10^{2n}}+\frac{3m}{10^{3m}}+\cdots. \end{array}$$

Since $3m<7m+1=10^n$, this gives an easy way to write out the repeating decimal expansion of the fraction $3/7$. In particular, $10^6-1=7\cdot142857$, and $3\cdot142857=428571$ so $3/7=0.\overline{428571}$.

This can be employed more generally. Without much loss of generality assume $0

$$\frac{a}{b}=\frac{am}{10^n}+\frac{am}{10^{2n}}+\frac{am}{10^{3m}}+\cdots.$$

Note again $am

Finally, for base $r$ (assuming $r$ is a positive natural number anyway), find $n$ such that $b|(r^n-1)$ and adapt this method. (Remember to pull out $\gcd(b,r)$ from the denominator.) Note that $n$ is guaranteed to exist because $r$ has an order modulo $b'$ ($b'$ coprime) by elementary number theory.

Also, if $d|r$, then $\frac{1}{d}=\frac{(r/d)}{r}$ is an easy way to compute the base-$r$ expansion of $1/d$. This helps with the "pulled-out" factors (i.e. with $d=\gcd(b,r)$).

10

Using the formula for the sum of a geometric series, we get $$ \sum_{k=1}^\infty\frac{3}{8^k}=\frac38\sum_{k=0}^\infty\frac{1}{8^k}=\frac38\frac{1}{1-\frac18}=\frac38\frac87=\frac37 $$ Therefore, $$ \frac37=\sum_{k=1}^\infty\frac{3}{8^k} $$

  • 0
    +1 As an exercise the OP should now write the fraction $3/7$ in base two.2012-08-02
5

Infinite series expansion for a number? There are many series whose is $\,3/7\,$, some of them rather dandy: $$\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(-6)^n}$$$$\frac{18}{7\pi^2}\sum^\infty_{n=1}\frac{1}{n^2}$$ and a long etc.

  • 0
    $\frac{1}{2} \sum_{n=0}^\infty \frac{1}{(-12)^n} = \frac{6}{13}$...2012-05-31
  • 0
    Sure, thanx. Corrected.2012-05-31
2

How about a few telescoping series?

$$\begin{align*} \frac37&=\sum_{k=1}^\infty\frac{40}{(7k+21)(3k+15)}\\ &=\sum_{k=1}^\infty\frac{100}{(17k+34)(3k+21)}\\ &=\sum_{k=1}^\infty\frac{36}{(k+5)(k+6)(k+7)}\\ &=\sum_{k=1}^\infty\frac{48}{(k+6)(k+7)(k+8)} \end{align*}$$

  • 0
    Maybe you can generalise your result [here](http://math.stackexchange.com/q/177932/19341)...?2012-08-02