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What is some infinite series expansion for 3/7? and so on for these fraction with digit in base 10? I can't think of some useful thing at all. Please generalize some useful series expression for all those kinds of fraction.

Can someone help?

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I don't understand the question. Long division already provides such an expansion. – Qiaochu Yuan May 30 '12 at 22:01
$0.428571428571428571428571\dots$. But you know this. What is the question? – André Nicolas May 30 '12 at 22:04
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. – Najib Idrissi May 31 '12 at 10:00
$\frac{3}{7}= \frac{3}{7} + 0 +0+0+ ..$ – N. S. May 31 '12 at 10:17
up vote 19 down vote accepted

One might also try the mundane

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

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Which now that I look again is rather diabolic... – Pedro Tamaroff May 30 '12 at 23:07

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 $\dfrac37$.

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I think Marvis wins the cake... – Willie Wong May 31 '12 at 9:17

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 $0<a<b$ with $b$ divisible by neither $2$ nor $5$ and find an $n$ such that $b|(10^n-1)$, in particular $10^n-1=bm$, so that


Note again $am<bm+1=10^n$ so this provides a clean decimal expansion. We ignore integers, and for rationals more generally we can decompose them into a sum of the nearest integer plus the rational's fractional part, and apply this method to the fractional part, then recombine. For rationals with $2,5$'s in the denominator's prime factorization, pull them out and multiply by them afterwards: their reciprocals are simply $0.5$ and $0.2$ after all.

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 $d=\gcd(b,r)$).

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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, $$ \frac37=\sum_{k=1}^\infty\frac{3}{8^k} $$

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+1 As an exercise the OP should now write the fraction $3/7$ in base two. – Jyrki Lahtonen Aug 2 '12 at 7:03

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.

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$\frac{1}{2} \sum_{n=0}^\infty \frac{1}{(-12)^n} = \frac{6}{13}$... – Brandon Carter May 31 '12 at 9:26
Sure, thanx. Corrected. – DonAntonio May 31 '12 at 10:11

How about a few telescoping series?

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

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Maybe you can generalise your result here...? – draks ... Aug 2 '12 at 9:28

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